Category: Class 11 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “s-Block Elements – Group 2 Elements : Alkaline Earth Metals”.

1. Which of the following is not an alkaline earth metal?
a) beryllium
b) boron
c) aluminium
d) calcium
Answer: a
Clarification: Beryllium, belongs to the 2^nd group. It is not called as an alkaline earth metal because the first element of this group that is beryllium is different from the rest of the members and it also shows a diagonal relationship with aluminium.

2. How many electrons do group 2 elements have in their S orbital of the valence shell?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: The alkaline earth metals have two electrons in the s-orbital of their valence shell and their general electronic configuration as represented as ns². Similar to the alkali metals these elements compounds are predominantly ionic.

3. Which of the following order is correct with respect to the hydration enthalpy?
a) B > Mg > Ca < Sr > Ba
b) Be⁺² > Mg⁺² > Ca⁺² > Sr⁺² > Ba⁺²
c) B > Mg < Ca > Sr > Ba
d) B > Mg > Ca > Sr < Ba
Answer: b
Clarification: The hydration enthalpy decreases with the increase in ionic size along with the group towards down, the correct order of hydration enthalpy is given as Be⁺² > Mg⁺² > Ca⁺² > Sr⁺² > Ba⁺² and the hydration enthalpies of alkaline earth metal ions are greater than the size of alkali metal ions. We can say that these are extensively hydrated than them.

4. What is the colour of barium?
a) brick red
b) crimson
c) apple green
d) blue
Answer: c
Clarification: The colours of Barium, Strontium and Calcium are Apple Green, Crimson and brick red. These are the colours of their flames, the colours occur when an electron is excited and jumps into a higher energy level and then drop back. They emit the radiation in the form of visible light.

5. Powdered beryllium burns in order to give ___________
a) barium sulphate
b) beryllium chloride
c) beryllium nitride
d) beryllium hydride
Answer: c
Clarification: We know that beryllium is kinetically inert to Oxygen and water as it forms oxide film on the surface. But whereas powdered beryllium burns brightly on ignition in the air in order to give the oxides and nitrides of beryllium.

6. Which of the following is the best route to prepare BeF₂?
a) thermal decomposition of BeF₂
b) thermal decomposition of beryllium
c) thermal decomposition of (NH₄)₂BeF₄
d) thermal decomposition of Barium sulphate
Answer: c
Clarification: All the alkali Earth metals combine with halogen at elevated temperatures forming their halides, the thermal decomposition of (NH₄)₂BeF₄, is best road and to prepare BeF₂ and BeCl₂ is conveniently made from the oxide.

7. Can beryllium hydride be prepared by combining with hydrogen through Heating?
a) Yes
b) No
c) Maybe
d) May not be
Answer: b
Clarification: All the alkaline earth metals (except beryllium) their hydrides can be prepared by combining with hydrogen. We can prepare beryllium hydride, through heating beryllium chloride with lithium aluminium hydride.

8. The reduction potential of alkaline earth metals is ___________ alkali metals.
a) may be equal to
b) greater than
c) less than
d) equal to
Answer: c
Clarification: Although the alkaline earth metals are reductants in nature, their reducing potential is not as greater as alkali metals. Beryllium has a negative value when compared to the other elements in its group, this is due to the large hydration energy associated with the small size of beryllium.

9. Which of the following is a component of milk of magnesia?
a) magnesium oxide
b) magnesium sulphate
c) magnesium hydroxide
d) magnesium chloride
Answer: c
Clarification: Suspension of magnesium Hydroxide in water is known as milk of magnesia and it is used as an antacid in order to treat acidity. It works as a strong base while magnesium carbonate is an ingredient and toothpaste.

10. With which of the following elements magnesium does not form an alloy?
a) manganese
b) aluminium
c) zinc
d) barium
Answer: d
Clarification: Magnesium combines with few elements like aluminium, manganese, zinc and tin in order to form a lot of magnesium aluminium alloys which are very light in mass and are used in the construction of aircraft. There are also many other uses of magnesium.

Chemistry Multiple Choice Questions on “Qualitative Analysis of Organic Compounds”.

1. Carbon and hydrogen are detected by heating the compound with which of the following?
a) Copper (II)oxide
b) Iron(II)oxide
c) Iron(III)oxide
d) Copper(I)oxide
Answer: a
Clarification: Carbon and hydrogen are detected by heating the compound with copper (II) oxide. In this way, the carbon that is present in the compound will be oxidized to carbon dioxide and hydrogen to water. Carbon dioxide is tested by limewater which develops turbidity, and hydrogen is tested with anhydrous copper sulphate, which turns blue.

2. Which compound gets precipitated in the detection of carbon and hydrogen?
a) Copper
b) Carbon dioxide
c) Calcium carbonate
d) Copper sulphate
Answer: c
Clarification: Calcium carbonate (CaCO₃) gets precipitated during the detection of carbon and hydrogen. Firstly carbon will react with copper (II) oxide to form copper and carbon dioxide. Then, hydrogen will react with the same copper (II) oxide to form copper and water. After this, the presence of carbon dioxide is tested by reacting it with calcium hydroxide and the product obtained is calcium carbonate, which gets precipitated and water.

3. Identify the element that cannot be detected by Lassaigne’s test.
a) Nitrogen
b) Fluorine
c) Sulfur
d) Phosphorous
Answer: b
Clarification: Fluorine cannot be detected by lassaigne’s test. Even though lassaigne’s test is used for the detection of halogens, all halogens except fluorine can be detected. This is because, in lassaigne’s test, the sodium extract is treated with silver nitrate. Only in the case of fluorine, the silver fluoride formed is soluble, unlike the others which is insoluble. Thus, the precipitate will not be formed and so, this method cannot be used for fluorine detection.

4. Potassium can replace sodium in lassaigne’s test.
a) True
b) False
Answer: a
Clarification: Potassium, like sodium is electropositive in nature. In lassaigne’s test, the elements present in the compound are converted from their covalent form to their ionic form by fusing the compound with sodium metal. Since, potassium has similar characteristics of electro positivity as sodium and since potassium is highly reactive, it can be used instead of sodium in lassaigne’s test.

5. What is Lassaigne’s test extract called as?
a) Fusion extract
b) Sodium fusion extract
c) Lassaigne extract
d) Sodium extract
Answer: b
Clarification: Lassaigne’s test extract is called as sodium fusion extract. The cyanides, sulphides and halides of sodium will be formed. These will be extracted from the fused mass by boiling it with distilled water. Hence, the name of the extract is sodium fusion extract.

6. In the test for nitrogen, the sodium fusion extract is acidified with which of the following?
a) Dilute sulphuric acid
b) Dilute hydrochloric acid
c) Concentrated hydrochloric acid
d) Concentrated sulphuric acid
Answer: d
Clarification: In the test for nitrogen, sodium cyanide first reacts with iron (III) sulphate and forms sodium hexacyanoferrate (II). On heating with concentrated sulphuric acid, some iron (II) ions are oxidized to iron (III)ions which react with sodium hexacyanoferrate (II) to produce iron (III) hexacyanoferrate (II), which is Prussian blue in color. Moreover, nitrogen atoms are soluble in concentrated sulphuric acid.

7. What is the color of the precipitate obtained in the test for sulphur?
a) White
b) Black
c) Violent
d) Blue
Answer: b
Clarification: In the test for sulphur, the sodium fusion extract is acidified with acetic acid and lead acetate is added to it. Once this reaction takes place, a black precipitate is formed. This black precipitate is lead acetate, indicating the presence of sulphur.

8. In case of both nitrogen and sulphur existence, Prussian blue is still the color of the end product.
a) True
b) False
Answer: b
Clarification: In case, both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed. The color formed is blood red and not Prussian blue. This is because, in this case, there are no free cyanide ions.
Na + C + N+ S → NaSCN
Fe³⁺ + SCN^– → [Fe (SCN)] ²⁺ (blood red)

9. A X color precipitate, which is Y in ammonium hydroxide indicates presence of chlorine. Identify X and Y.
a) X = yellowish, Y = soluble
b) X = yellow, Y = insoluble
c) X = white, Y = insoluble
d) X = white, Y = soluble
Answer: d
Clarification: During the detection of chlorine, when the organic compound reacts with sodium, it forms sodium chloride. This sodium chloride gives the white precipitate of silver nitrate with silver nitrate solution. This white precipitate is also soluble in ammonium hydroxide.

10. Which is the oxidizing agent used in the test for phosphorous?
a) Hydrogen peroxide
b) Potassium nitrate
c) Sodium peroxide
d) Nitric acid
Answer: c
Clarification: The phosphorous present in the organic compound is oxidized to phosphate by the oxidizing agent sodium peroxide. This solution is then boiled with nitric acid and treated with ammonium molybdate. As a result, a yellow precipitate is formed, indicating the presence of phosphorous.

Chemistry Questions and Answers for Aptitude test on “Uncertainty in Measurement”.

1. Write 6354000000 in scientific notation.
a) 6.354 x 10⁹
b) 6354 x 10⁶
c) 0.64 x 10¹⁰
d) 6354000 x 10³
Answer: a
Clarification: Scientific notation should have the least possible significant figures as the zeroes are made as to the power of ten. Or simply the power of ten in scientific notation is equal to the number of times the decimal point moved to produce a number between 1 and 10.

2. _________ is referred to as the closeness of different measurements for the same quantity.
a) Accuracy
b) Precision
c) Analysis
d) Dimension
Answer: b
Clarification: Precision is referred to as the closeness of different measurements for the same quantity. Accuracy is the degree to which it’s taken as standard or almost equivalent to it.

3. How many seconds are there in a half day?
a) 86,400 seconds
b) 43,200 seconds
c) 172,800 seconds
d) 3660 seconds
Answer: b
Clarification: Half day means 12 hours. One hour means 60 seconds. And each minute is of 60 seconds. In the above question, we need the total number of seconds so 12 x 60 x 60 seconds = 43,200 seconds.

4. A piece of iron is 5 inches long. How much would it be in centimeters?
a) 12.7 cm
b) 6.35 cm
c) 5 cm
d) 500 cm
Answer: a
Clarification: 1 inch = 2.54 cm : 1 cm = 0.3931 inch. As we now know, how much is 1inch in centimeters, then we need to multiply 5 to 2.54 cm in order to convert it into centimeters. 5 x 2.54 cm = 12.7 cm.

5. How many significant figures does 0.057 have?
a) 2
b) 4
c) 3
d) 0
Answer: a
Clarification: The non-zero digits and any zeros between them are all significant. Leading zeros are not significant. Counting all the significant digits gives us 2.

6. How many significant figures does 63180 have?
a) 5
b) 4
c) 1
d) 2
Answer: b
Clarification: The non-zero digits and any zeros between them are all significant. Since there is no decimal, zeros are not significant. Therefore it’s 4.

7. The exact value is 150m. A students record it as 140.1m in 1st turn and 140.8m in the 2^nd turn. Comment his/her recordings.
a) precise
b) accurate
c) neither precise nor accurate
d) both precise and accurate
Answer: a
Clarification: These values are precise as they are close to each other but are not accurate. Precision is referred to as the closeness of different measurements for the same quantity. Accuracy is the degree to which it’s taken as standard or almost equivalent to it.

8. The exact value is 150m. A students record it as 149.1m in 1st turn and 150.8m in the 2^nd turn. Comment his/her recordings.
a) precise
b) accurate
c) neither precise nor accurate
d) both precise and accurate
Answer: b
Clarification: These values are accurate as they are nearer to the exact value. Precision is referred to as the closeness of different measurements for the same quantity. Accuracy is the degree to which it’s taken as standard or almost equivalent to it.

9. Multiply 1.2 and 3.91. Obtain the result as per the rules of significant figures.
a) 4.692
b) 4.69
c) 5
d) 4.7
Answer: d
Clarification: 1.2 x 3.91 = 4.692. Since 1.2 has two significant figures, the result should not have more than two significant figures, thus, it is 4.7. As per the rules of significant figures, the resultant answer should not have more significant figures the number with a less significant figured number.

10. How many significant figures are there in 60.6?
a) 4
b) 2
c) 3
d) 1
Answer: c
Clarification: The non-zero digits and any zeros between them are all significant. Leading zeros are not significant. Counting all the significant digits gives us 3.

Chemistry Multiple Choice Questions on “Electronic Configurations of Elements and the Periodic Table”.

1. The electrons’ distribution into the atomic orbitals is called as _________
a) Electronic order
b) Electronic distribution
c) Electronic filing
d) Electronic configuration
Answer: d
Clarification: When electrons are distribution into the orbitals of the atoms by following Aufbau’s principal, Hund’s rule maximum multiplicity and Pauli’s exclusive principle, the order they are filled is represented by the electronic configuration.

2. The location of any element in the periodic table is determined by the __________ of the filled last orbital.
a) Spin quantum number
b) Quantum numbers
c) Azimuthal quantum number
d) Magnetic quantum number
Answer: b
Clarification: Quantum numbers are responsible for an element’s position in the periodic table. For example, as we know the principal quantum number defines the main energy level i.e. shell. The last orbital principal quantum number is the periodic number of the element.

3. How many electrons are there in the L-shell of Boron?
a) 3
b) 2
c) 1
d) 4
Answer: a
Clarification: The atomic number of Boron is 5 and it contains two elements in the K-shell and three elements in the L-shell as well as 4 electrons in s-orbital and 1 electron and p-orbital. The element preceding Boron is beryllium and the element succeeding Boron is carbon.

4. What are the d-block elements called?
a) Inner transition elements
b) Alkali earth metals
c) Transition elements
d) Noble gases
Answer: c
Clarification: The d-block elements are called transition elements as the electrons are energetically filled. The exact meaning of transition elements is that elements have either partially or fully filled subshell or can actively cations by losing electrons.

5. 3d transition series starts from _________ and ends with __________
a) Zinc, Scandium
b) Scandium, Zinc
c) Vanadium, Nickel
d) Argon, Zinc
Answer: b
Clarification: In the 3d transition series, the have either partially or fully filled 3d orbitals. The 3d transition state starts from Scandium (Z = 21) having electronic configuration as 3s²3d¹ and ends with Zinc (Z = 30) whose electronic configuration is 3s²3d¹⁰.

6. Which block elements are known as inner transition elements?
a) s-block
b) p-block
c) d-block
d) f-block
Answer: d
Clarification: Inner transition elements are f-block i.e. 4f and 5f. They include elements from 57 to 71 and 89 to 103. The 4f elements are lanthanoids (57 – 71) and 5f are actinides (89 – 113). Most of the actinides are radioactive.

7. The Lanthanoid series is the same as the Actinoid series.
a) True
b) False
Answer: b
Clarification: Both the Lanthanoid series and the Actinoid series make up Inner transition elements, but they are not the same. The 4f elements are lanthanoids (Z = 57 to Z = 71) and 5f are actinides (Z = 89 to Z = 113).

8. Write the correct electronic configuration of Neon.
a) 1s²2s²2p⁶
b) 1s²2s²
c) 1s²2p⁶
d) 2s²2p⁶
Answer: a
Clarification: Neon is made up of 10 electrons. As per Aufbau’s principle, orbitals with lower energy must be filled first and s-orbital can only accommodate 2 electrons while p can accommodate 6 electrons. So the electronic configuration of Neon is 1s²2s²2p⁶.

9. Group I elements are called as ________
a) Alkali metals
b) Noble gases
c) Chalcogens
d) Halogens
Answer: a
Clarification: Group I elements are known as Alkali metals as they have a great ability to form strong bases i.e. alkalies. Alkalies are the strong bases with the ability to neutralize strong acids. They consists of Lithium(Li), Sodium(Na), Potassium(K), Rubidium(Rb), Cesium(Cs) and Francium(Fr).

10. The fourth period ends with the element ________
a) Helium
b) Argon
c) Neon
d) Krypton
Answer: d
Clarification: Every period ends with a noble gas, as the element at the end of the period is with the fully filled orbitals. The 4^th period has an outer shell as O i.e. n = 4. So the configuration is [Ar}4s²3d¹⁰4p⁶, that is Krypton.

Chemistry online quiz on “States of Matter – Gas Laws”.

1. At a constant temperature, the pressure of a gas is given as one atmospheric pressure and 5 liters. When the atmospheric pressure is increased to 2 atm, then what is the volume of the gas?
a) 1 liter
b) 5 liters
c) 10 liters
d) 2.5 liters
Answer: d
Clarification: As we know that, Boyle’s law states at a constant temperature, the pressure of a gas is inversely proportional to its volume so P₁V₁ equals to P₂V₂ by substituting P₁ as one atmospheric pressure V₁ as 5 liters P₁ as to atmospheric pressure we get V₂ as 5/2 that is 2.5 liters.

2. What is the shape of the graph that is drawn between pressure and volume?
a) A straight line
b) Circular
c) Parabola
d) Hyperbola
Answer: d
Clarification: Boyle’s law states that at constant temperature pressure is inversely proportional to the volume of gas so here the graph between the pressure as y-axis volume as x-axis is in the shape of a hyperbola.

3. What is the name of the graph that is drawn, when the temperature is kept constant?
a) Isotherm
b) Isochoric and isobar
c) Isochoric
d) Isobar
Answer: a
Clarification: The graphs with constant temperature plot are isotherms. For example, the graph that is used to detect the Boyle’s law, that is between pressure and volume is an isotherm as the temperature is constant in this graph.

4. There is a ball that will burst if the pressure exceeds 0.12 bars. The pressure of the gas is 1 bar and the volume is 2.5 liters. What can be the maximum volume that the ball can be expanded?
a) 0.12 liters
b) 2.5 liters
c) 0.3 liters
d) 1 liter
Answer: c
Clarification: According to Boyle’s law at a constant temperature, the pressure is inversely proportional to the temperature so here P₁V₁ is equaled to P₁V₂ by equating P₁V₁ is equaled to 1 x 2.5 = 2.5, so the maximum volume of the ball that can be expanded is 2.5/0.12 =0.3 liters.

5. How much does the volume of the gas increase if we increase the temperature by 1 Degree?
a) 273 liters
b) 1 by 273^rd of the original volume of the gas
c) 1 liter
d) Hundred liters
Answer: b
Clarification: According to Charles law, the volume of the fixed gas at constant pressure is directly proportional to the Absolute Temperature of the gas. So we thereby represent this as V_t = V₀(1 + t/273) where V_t is the volume of the gas at temperature t and V₀ is the volume of the gas at 0 degrees Celsius that is absolute temperature.

6. There is a balloon filled with a gas at 26-degree centigrade and has a volume of about 2 liters when the balloon is taken to a place which is at 39-degree centigrade, what would be the volume of the gas that is inside the balloon?
a) 2 liters
b) 3 liters
c) 1.5 liters
d) 0.67 liters
Answer: b
Clarification: As we know that temperature is directly proportional to the volume at constant pressure, 26/39 = 2/ X; so here by equating X equals to 3 liters. Hence required a volume of the balloon at 39 degrees is 3 liters.

7. By observing the below-given figure which of the options do you think is the correct one?

a) P₁ is greater than P₂
b) P₂ is greater than P₁
c) P₁ is equal to P₂
d) P₁ may be equal to P₂
Answer: a
Clarification: By using Boyle’s law, draw a parallel line to volume axis, so as to maintain a constant temperature. Draw perpendicular lines to the point of intersection of pressure lines to constant temperature and volume axis. Now see to the lower the volume, higher the pressure. So P₁ is greater than P₂.

8. An ideal gas of 10 moles occupies _________ volume.
a) 22.4 liters
b) 2.24 liters
c) 224 liters
d) 2240
Answer: c
Clarification: As we know that the number of moles is proportional to the volume as per Avogadro’s law and we also know that an ideal gas at STP occupies 22.4 liters of volume. So here 10 moles of gas occupies 224 liters of volume.

9. When a graph is drawn between the pressure and temperature of the gas it is known as _________
a) isochoric
b) isobar
c) isotherm
d) isotopic
Answer: a
Clarification: When a graph is plotted between pressure on y-axis and temperature on x-axis straight line is formed at a constant volume and this graph is known as isochoric. As we know that gay lussac’s law proposes that at the constant volume the pressure and temperature are directly proportional.

10. At 22 degree Celsius a gas consists of pressure 1.1 bars then what is the temperature when the gas consists a pressure of 2.2 bars?
a) 11 degree Celsius
b) 44 degree Celsius
c) 33 degree Celsius
d) 22 degree Celsius
Answer: b
Clarification: According to Gay-Lussac’s law, at the constant volume, the pressure is directly proportional to the temperature of gas so P₁/P₂ = T₁/T₂ that is 1.1 Bar/2.2 bar = 22 degrees Celsius/44 degree Celsius. So the temperature required is 44 degrees Celsius.

To practice all areas of Chemistry for online Quizzes,

Chemistry Multiple Choice Questions on “Thermodynamics – Measurement of ∆U and ∆H: Calorimetry”.

1. Bomb calorimeter is __________ in nature.
a) isothermal
b) isochoric
c) isobaric
d) absolute
Answer: b
Clarification: ABomb calorimeter works at constant volume, so it is isochoric in nature. Here, the heat energy that’s measured is only the internal energy. The work done is zero because of no change in volume.

2. If an exothermic reaction occurs in a Bomb calorimeter then the temperature of the water bath ________
a) increases
b) decreases
c) remains constant
d) cannot predict
Answer: a
Clarification: In a Bomb calorimeter, the reaction occurs in a vessel and that is surrounded by the water bath. If it is an exothermic reaction the temperature rises and if it is an endothermic reaction the temperature decreases. Temperature can be measured using a thermometer.

3. Heat capacity of a Bomb calorimeter is given by _______
a) C_V
b) C_P
c) C_M
d) C_B
Answer: a
Clarification: A bomb calorimeter operates at a constant volume i.e. it is an isochoric process. So the heat capacity is C_V which represents heat capacity at constant volume. While C_P represents heat capacity at constant pressure.

4. Bomb calorimeter is used to determine ____________
a) molar heat capacity
b) heat of combustion
c) rate kinetics
d) affinity
Answer: b
Clarification: A bomb calorimeter is used to measure the heat of combustion of a reaction. It has to withstand a large amount of pressure, in order to determine the heat of combustion. It is an isochoric process and the heat energy is equal to the internal energy.

5. “c” the specific heat capacity of a substance is given by temperature difference is given by ΔT and the heat energy is given by Q then what is the mass of the substance?
a) Q/cΔT
b) cΔT /Q
c) QcΔT
d) QΔT
Answer: a
Clarification: Heat energy of a substance is denoted by Q and is given by the expression mcΔT, where m is a mass of the substance, c is a specific heat capacity and ΔT is temperature difference, so the mass of the substance is Q/cΔT.

6. During the process of conversion of ice into the water, the specific heat capacity is given by _______
a) 0
b) positive
c) infinity
d) negative
Answer: c
Clarification: During the phase change of a substance the temperature change is zero. As we know that specific heat capacity = Q/cΔT which is zero; specific heat capacity becomes infinity, so during the process of conversion of ice into the water, the specific heat capacity is infinity.

7. What is the value of specific heat capacity in the adiabatic process?
a) 0
b) infinity
c) positive
d) negative
Answer: a
Clarification: During an adiabatic process the change in total energy is zero, As we know that the specific heat capacity is given by the total heat required by mass X change in temperature, heat energy is zero the specific heat capacity becomes zero.

8. When 1 kg of water at 373 k, is converted into steam how much amount of heat energy is required?
a) 22600 KJ
b) 226 KJ
c) 2260 KJ
d) 22.6 KJ
Answer: c
Clarification: The latent heat of vaporization of water is 2260 KJ/Kg. The heat that is required to convert water at 373 k to steam is given by Q = mL, where m = mass of the water and L = latent heat vaporization of water; heat energy required = 1 kg x 2260 KJ/Kg = 2260 KJ.

9. The total heat energy utilized for increasing the temperature by 4 degrees Kelvin in a 3 kgs substance is 100 KJ what is the specific heat capacity of that substance?
a) 8.34 KJ/g-k
b) 8.34 KJ/Kg-k
c) 8.34 KJKg-k
d) 8.34 KJ/Kg
Answer: b
Clarification: The formula of heat energy is given by the expression: Q = mcΔT, m is the mass of the substance, c is the specific heat capacity and ΔT is the temperature difference. c = Q/mΔT; c = 100KJ/3kg(4k) = 8.34 KJ/Kg-k.

10. The enthalpy, internal energy during a process and change in volume are 500 units, 400 units, and 2 units. What is the pressure that is exerted on the gas during this process?
a) 20 units
b) 80 units
c) 100 units
d) 50 units
Answer: d
Clarification: We know that ΔH = ΔU + PΔV; ΔH is the enthalpy, ΔU is the internal energy, ΔV is the change in volume and P is the pressure. So by substituting the enthalpy, internal energy during a process and change in volume as 500 units, 400 units and 2 units, we get pressure as 50 units.