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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.

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