## Chapter 3

## Q. 3.4

A gas in its ideal-gas state undergoes the following sequence of mechanically reversible

processes in a closed system:

(a) From an initial state of 70°C and 1 bar, it is compressed adiabatically to 150°C.

(b) It is then cooled from 150 to 70°C at constant pressure.

(c) Finally, it expands isothermally to its original state.

Calculate W, Q, ΔU^{ig}, and ΔH^{ig} for each of the three processes and for the entire cycle.

Take C^{ig}_{V}=12.471 and C^{ig}_{P} = 20.785 J· mol^{−1}· K^{−1}.

## Step-by-Step

## Verified Solution

Take as a basis 1 mol of gas.

(a) For adiabatic compression, Q = 0, and

Δ U^{ig} = W = C^{ig}_{V} ΔT = ( 12.471 ) ( 150 − 70 ) = 998 J

Δ H^{ig} = C^{ig}_{P} ΔT = ( 20.785 ) ( 150 − 70 ) = 1663 J

Pressure P_{2} is found from Eq. (3.23b):

TP^{(1 -\gamma )/\gamma } = const (3.23b)

P_{2} = P_{1} \left(\frac{T_{2} }{T_{1}}\right) ^{\gamma (\gamma – 1)} = (1)\left(\frac{150 + 273.13}{70 + 273.15} \right) ^{2.5} = 1.689 bar

(b) For this constant-pressure process,

Q = Δ H^{ig} = C _{P}^{ig} ΔT = ( 20.785 ) ( 70 − 150 ) = −1663 J

ΔU = C _{V}^{ig} ΔT = ( 12.471 ) ( 70 − 150 ) = −998 J

W = Δ U^{ig} − Q = −998 − ( −1663 ) = 665 J

(c) For this isothermal process,ΔU^{ig} and ΔH^{ig} are zero; Eq. (3.20) yields:

\varrho = – W RT In \frac{V^{ig}_{2} }{V^{ig}_{1} } = RT In \frac{P_{1} }{P_{2}} ( const T ) (3.20)

\varrho = -W = RT In \frac{P_{3}}{P_{1}} = RT In \frac{P_{2}}{P_{1}} = ( 8.314 )( 343.15 ) ln\frac{1.689}{1} = 1495 J

For the entire cycle,

Q = 0 − 1663 + 1495 = −168 J

W = 998 + 665 − 1495 = 168 J

Δ U^{ig} = 998 − 998 + 0 = 0

Δ H^{ig} = 1663 − 1663 + 0 = 0

The property changes ΔU^{ig} and ΔH^{ig} both are zero for the entire cycle because the initial and final states are identical. Note also that Q = −W for the cycle. This follows from the first law with ΔU^{ig} = 0.