Answer to Question #244319 in Molecular Physics | Thermodynamics for Zyy

Question

Heat is transferred to 6 kg of air which is initially at 100 kPa and B
0C until its temperature reaches 600
K and pressure reaches at 150 kPa. Determine the change in internal energy, the change in enthalpy,
the heat supplied and the work done in the following processes:
1. Constant Volume Process followed by Constant PressureProcess
2. Constant Pressure process followed by Constant VolumeProcess
Assume that air is an ideal gas. R=8.314 KJ/Kmol K, Cp=20.785 KJ/Kmol K and Cv =20.785
KJ/Kmol KMolecular weight of air =29. B = 42

Answer 1

1. Constant Volume Process followed by Constant Pressure Process: At first, we have:

$n=\frac{6\,kg}{0.029\,kg/mol}=206.897\,mol\,of\,air \\ \text{ } \\ T_1=42\,°C=315\,K \\ V_1=\frac{nRT_1}{P_1} \\ V_1=\cfrac{(206.897\,mol)(8.314\frac{kPaL}{mol\,K})(315\,K)}{100\,kPa} \\ V_1=5418.446\,L$

First, since V₁=V₂ (for the constant volume process):

$T_2=T_1\frac{P_2}{P_1}=(315\,K)(\frac{150\,kPa}{100\,kPa})=472.5\,K$

Then, we know that for an isochoric process $W=\intop pdV=0$, this implies that the first law of thermodynamics is now $\Delta U = Q-W=Q=nC_v(T_2-T_1)$ and for the enthalpy change $\Delta H = nC_p(T_2-T_1)$.

$W=0\,J \\ \Delta U =Q=(206.897\,mol)(20.785\frac{J}{mol\,K})(472.5-315)K \\ \Delta U =Q= 677305.778\,J \\ \Delta H=(206.897\,mol)(20.785\frac{J}{mol\,K})(472.5-315)K \\ \Delta H= 677305.778\,J$

The second part is an isobaric process, where

$P_2=P_3=150\,kPa \\ T_2=472.5\,K; T_3=600\,K \\ V_2=5418.446\,L \\ V_3=V_2\frac{T_3}{T_2}=(5418.446\,L)(\frac{600\,K}{472.5\,K}) \\ V_3=6880.566\,L$

Now we can find the rest of the terms knowing the same equations as before for $\Delta U$ and $\Delta H$:

$W=P_2(V_3-V_2) \\ W=(150\,kPa)(6880.566-5418.446)L=219318\,J \\ \Delta U =(206.897\,mol)(20.785\frac{J}{mol\,K})(600-472.5)K \\ \Delta U =548295.153\,J \\ Q=\Delta U + W = 767613.153\,J \\ \Delta H=(206.897\,mol)(20.785\frac{J}{mol\,K})(600-472.5)K \\ \Delta H= 548295.153\,J$

The conclusion for the first part is

$W=(0+219318)\,J=219318\,J=219.318\,kJ \\ Q=(677305.778+767613.153)\,J=1444918.931\,J=1444.918\,kJ \\ \Delta U=(677305.778+548295.153)\,J=1225600.931 \,J=1225.601\,kJ \\ \Delta H=(677305.778+548295.153)\,J=1225600.931 \,J=1225.601\,kJ$

2. Constant Pressure process followed by Constant Volume Process

First, we know that

$n=\frac{6\,kg}{0.029\,kg/mol}=206.897\,mol\,of\,air \\ \text{ } \\ T_1=42\,°C=315\,K \\ V_1=\frac{nRT_1}{P_1} \\ V_1=\cfrac{(206.897\,mol)(8.314\frac{kPaL}{mol\,K})(315\,K)}{100\,kPa} \\ V_1=5418.446\,L$

First, since P₁=P₂ (for the constant pressure process):

$V_2=V_1\frac{T_2}{T_1}=(5418.446\,L)(\frac{600\,K}{315\,K})=10320.849\,L$

Now, we know that for this process we can calculate the following:

$W=P_1(V_2-V_1)=(100\,kPa)(10320.849-5418.446)L=490240.3\,J \\ \Delta U =(206.897\,mol)(20.785\frac{J}{mol\,K})(600-315)K \\ \Delta U =1225600.931\,J \\ Q=\Delta U + W = 1775841.231\,J \\ \Delta H=(206.897\,mol)(20.785\frac{J}{mol\,K})(600-315)K \\ \Delta H= 1225600.931\,J$

Now, the second part of this other process involves a process at constant volume, where

$V_2=V_3=10320.849\,L \\ T_2=T_1\frac{P_2}{P_1}=(600\,K)(\frac{150\,kPa}{100\,kPa})=900\,K$

Then, since the volume is constant no work is done thus

$W=0\,J \\ \Delta U =Q=(206.897\,mol)(20.785\frac{J}{mol\,K})(900-600)K \\ \Delta U =Q= 1290106.244\,J \\ \Delta H=(206.897\,mol)(20.785\frac{J}{mol\,K})(900-600)K \\ \Delta H= 1290106.244\,J$

For the second part, we have the following

$W=(490240.3+0)\,J=490240.3\,J=490.240\,kJ \\ Q=(1775841.231+1290106.244)\,J=3065947.475\,J=3065.947\,kJ \\ \Delta U=(1225600.931+1290106.244)\,J=2515707.175\,J=2515.707\,kJ \\ \Delta H=(1225600.931+1290106.244)\,J=2515707.175\,J=2515.707\,kJ$

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.