Question 3.4: A gas in its ideal-gas state undergoes the following sequenc...

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 $ΔU^{ig} = 0$.