Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 12.6, Problem 40P

(a)

To determine

To drive an expression of Δu for a gas whose equation of state is P(va)=RT for an isothermal process.

(a)

Expert Solution
Check Mark

Answer to Problem 40P

The expression of Δu for a gas whose equation of state is P(va)=RT for an isothermal process is Δu=0.

Explanation of Solution

Write the equation of state of the given gas.

P(va)=RT (I)

Here, the temperature is T, the pressure is P, the specific volume is v, the universal gas constant is R and the constant is a.

Write the general expression for change in internal energy (Δu).

Δu=u2u1=T1T2cvdT+v1v2(T(PT)vP)dv (II)

Here, the internal energy at state 1, 2 is u1,u2 and the specific heat at constant volume is cv; the symbol indicates the partial derivative of variables.

Rearrange the Equation (I) to obtain P.

P=RTva (III)

Conclusion:

Partially differentiate the Equation (III) with respect to temperature by keeping the specific volume as constant.

(PT)v=T(RTva)v=Rva(TT)v=Rva(1)=Rva

Substitute Rva for (PT)v in Equation (II).

Δu=T1T2cvdT+v1v2(T(Rva)P)dv=T1T2cvdT+v1v2(RTvaP)dv=T1T2cvdT+v1v2(PP)dv=T1T2cvdT+v1v2(0)dv

=T1T2cvdT (IV)

For an isothermal process, the temperature is kept constant.

T=constant

The differential temperature or change in temperature becomes zero.

dT=0

Substitute 0 for dT in Equation (IV).

Δu=T1T2cv(0)=0

Thus, the expression of Δu for a gas whose equation of state is P(va)=RT for an isothermal process is Δu=0.

(b)

To determine

To drive an expression of Δh for a gas whose equation of state is P(va)=RT for an isothermal process.

(b)

Expert Solution
Check Mark

Answer to Problem 40P

The expression of Δh for a gas whose equation of state is P(va)=RT for an isothermal process is Δh=a(P2P1).

Explanation of Solution

Write the general expression for change in enthalpy (Δh).

Δh=h2h1=T1T2cpdT+P1P2(vT(vT)P)dP (V)

Here, the enthalpy at state 1, 2 is h1,h2 and the specific heat at constant pressure is cp; the symbol indicates the partial derivative of variables.

Rearrange the Equation (V) to obtain v.

v=RTP+a (VI)

Conclusion:

Partially differentiate the Equation (VI) with respect to temperature by keeping the pressure as constant.

(vT)P=T(RTP+a)P=T(RTP)P+T(a)P=RP(TT)P+0=RP(1)

=RP

Substitute RP for (vT)P in Equation (V).

Δh=T1T2cpdT+P1P2(vT(RP))dP=T1T2cpdT+P1P2(vRTP)dP=T1T2cpdT+P1P2[v(va)]dP=T1T2cpdT+P1P2[vv+a]dP

=T1T2cpdT+P1P2adP=T1T2cpdT+a(P2P1) (VI)

For an isothermal process, the temperature is kept constant.

T=constant

The differential temperature or change in temperature becomes zero.

dT=0

Substitute 0 for dT in Equation (VI).

Δh=T1T2cp(0)+a(P2P1)=0+a(P2P1)=a(P2P1)

Thus, the expression of Δh for a gas whose equation of state is P(va)=RT for an isothermal process is Δh=a(P2P1).

(c)

To determine

To drive an expression of Δs for a gas whose equation of state is P(va)=RT for an isothermal process.

(c)

Expert Solution
Check Mark

Answer to Problem 40P

The expression of Δs for a gas whose equation of state is P(va)=RT for an isothermal process is Δs=R(lnP2P1).

Explanation of Solution

Write the general expression for change in entropy (Δs).

Δs=s2s1=T1T2cpTdTP1P2(vT)PdP (VII)

Here, the entropy at state 1, and 2 are s1 and s2.

Conclusion:

Substitute RP for (vT)P in Equation (VII).

Δs=T1T2cpTdTP1P2(RP)dP=T1T2cpTdTRP1P2(1P)dP=T1T2cpTdTR(lnP2P1) (VIII)

For an isothermal process, the temperature is kept constant.

T=constant

The differential temperature or change in temperature becomes zero.

dT=0

Substitute 0 for dT in Equation (VIII).

Δs=T1T2cpT(0)R(lnP2P1)=0R(lnP2P1)=R(lnP2P1)

Thus, the expression of Δs for a gas whose equation of state is P(va)=RT for an isothermal process is Δs=R(lnP2P1).

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Chapter 12 Solutions

Thermodynamics: An Engineering Approach

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