This set of Thermodynamics Problems focuses 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

View Answer

Explanation: 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

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Explanation: 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

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Explanation: 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

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Explanation: 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)

View Answer

Explanation: 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

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Explanation: 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

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Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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 .

**Sanfoundry Global Education & Learning Series – Thermodynamics.**

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