Thermodynamics Problems on “Maxwell’s Equations and TDS Equations”.
1. If a relation exists among variables x,y,z then z may be expressed as a function of x and y as, dz=Mdx+Ndy .
a) true
b) false
Answer: a
Clarification: Here, M,N and z are functions of x and y.
2. A pure substance which exists in a single phase has ____ independent variables.
a) zero
b) one
c) two
d) three
Answer: c
Clarification: Of all the quantities, any one can be expressed as a function of any two others.
3. Which of the following relation is correct?
a) dU=TdS-pdV
b) dH=TdS+Vdp
c) dG=Vdp-SdT
d) all of the mentioned
Answer: d
Clarification: These relations are true for a pure substance which undergoes an infinitesimal reversible process.
4. Maxwell’s equations consists of ____ equations.
a) four
b) three
c) two
d) one
Answer: a
Clarification: Maxwell’s equations consists of four equations.
5. Which of the following is not a Maxwell equation?
a) (∂T/∂V) = -(∂p/∂S)
b) (∂T/∂p) = -(∂V/∂S)
c) (∂p/∂T) = (∂S/∂V)
d) (∂V/∂T) = -(∂S/∂p)
Answer: b
Clarification: The correct equation is (∂T/∂p) = (∂V/∂S).
6. The condition for exact differential is
a) (∂N/∂y) = (∂M/∂x)
b) (∂M/∂y) = (∂N/∂x)
c) (∂M/∂y) = -(∂N/∂x)
d) all of the mentioned
Answer: b
Clarification: This is the condition for perfect or exact differential and here M and N are the functions of x and y.
7. The first TdS equation is
a) TdS=Cv*dT + T(∂T/∂p)dV
b) TdS=Cv*dT – T(∂p/∂T)dV
c) TdS=Cv*dT + T(∂p/∂T)dV
d) TdS=Cv*dT – T(∂T/∂p)dV
Answer: c
Clarification: This equation comes when entropy is defined as a function of T and V and using Cv and Maxwell’s third equation.
8. The second TdS equation is
a) TdS=Cp*dT + T(∂V/∂T)dp
b) TdS=Cp*dT – T(∂V/∂T)dp
c) TdS=Cp*dT + T(∂T/∂V)dp
d) TdS=Cp*dT – T(∂T/∂V)dp
Answer: b
Clarification: This equation comes when entropy is defined as a function of T and p and using Cp and Maxwell’s fourth equation.
9. Which of the following is true?
a) (∂p/∂V)*(∂V/∂T)*(∂T/∂p)= infinity
b) (∂p/∂V)*(∂V/∂T)*(∂T/∂p)= 0
c) (∂p/∂V)*(∂V/∂T)*(∂T/∂p)= 1
d) (∂p/∂V)*(∂V/∂T)*(∂T/∂p)= -1
Answer: d
Clarification: This is the relation between the thermodynamic variables, p,V and T.
10. For getting TdS equations, we assume entropy to be a function of T and V and also of T and p.
a) true
b) false
Answer: a
Clarification: For first TdS equation, we assume entropy as a function of T and V and for second TdS equation, we assume entropy as a function of T and p .
To practice problems on all areas of Thermodynamics,