Physics Multiple Choice Questions on “Thermodynamic Processes”.
1. In which of the following processes is heat transfer equal to zero?
a) Isentropic
b) Isochoric
c) Isothermal
d) Diathermic
Answer: a
Clarification: Entropy is defined as dQ/T. In an isentropic process dQ/T = 0 which means dQ = 0. Diathermic is a process in which heat flow is easily possible. Isochoric process is one in which volume is constant. Isothermal process is one in which temperature stays constant.
2. Isothermal process can be represented by which law?
a) Charle’s law
b) Boyle’s law
c) Gay-Lussac’s law
d) 2nd law of thermodynamics
Answer: b
Clarification: In an isothermal process, PV=const. This is the same as Boyle’s law. Charle’s law is given by: V/T=const. Gay-Lussac’s law is given by: P/T=const and 2nd law of thermodynamics states that in every process total entropy of the universe must increase.
3. In an isothermal process for an ideal gas, change in internal energy is zero. True or False?
a) True
b) False
Answer: a
Clarification: For an ideal gas internal energy depends only on temperature. In an isothermal process change in temperature is zero, hence internal energy also remains the same.
4. Calculate the work done by the gas in an isothermal process from A to B. PA = 1Pa, VA = 3m3, PB = 3Pa.
a) 3.3J
b) 3J
c) -3.3J
d) -4.58J
Answer: c
Clarification: Since the process is isothermal the product PV will be constant.
PAVA = PBVB.
∴ VB = 1*3/3 = 1m3.
Work done in an isothermal process is given by:
nRT*ln(VB/VA)
= PAVAln(VB/VA)
= 3ln(1/3)
= -3.3J.
5. Which of the following variables is zero for a cyclic process?
a) Work done
b) Heat supplied
c) Total heat + Total work
d) Total heat – Total work
Answer: d
Clarification: In a cyclic process, the starting position is the same as the ending position. So, the change in all state variables is zero. So the net change in internal energy is zero. ΔU = ΔQ – ΔW = 0.