To calculate the internal energy ($\Delta U$) and enthalpy ($\Delta H$) changes for air, given the conditions in the image, we'll use the following formulas for an ideal gas:

1. Change in internal energy: $\Delta U = n C_v \Delta T$ where:

   • $n$ is the number of moles (not directly given, but we assume conditions are per kmol),
   • $C_v$ is the specific heat at constant volume (21 kJ/kmol·K),
   • $\Delta T$ is the temperature change, $\Delta T = T_2 - T_1$.

2. Change in enthalpy: $\Delta H = n C_p \Delta T$ where:

   • $C_p$ is the specific heat at constant pressure (29.3 kJ/kmol·K),
   • $\Delta T = T_2 - T_1$.

Given dataTo calculate the internal energy ($\Delta U$) and enthalpy ($\Delta H$) changes for air, given the conditions in the image, we'll use the following formulas for an ideal gas:

1. Change in internal energy: $\Delta U = n C_v \Delta T$ where:

   • $n$ is the number of moles (not directly given, but we assume conditions are per kmol),
   • $C_v$ is the specific heat at constant volume (21 kJ/kmol·K),
   • $\Delta T$ is the temperature change, $\Delta T = T_2 - T_1$.

2. Change in enthalpy: $\Delta H = n C_p \Delta T$ where:

   • $C_p$ is the specific heat at constant pressure (29.3 kJ/kmol·K),
   • $\Delta T = T_2 - T_1$.

Given

• Initial temperature $T_1 = 277 \, \text{K}$,
• Final temperature $T_2 = 333 \, \text{K}$,
• Initial pressure $P_1 = 10 \, \text{bar}$,
• Final pressure $P_2 = 1 \, \text{atm} \approx 1.013 \, \text{bar}$,
• Initial molar volume $V_m = 2.28 \, \text{m}^3/\text{kmol}$,
• $C_v = 21 \, \text{kJ/kmol.K}$,
• $C_p = 29.3 \, \text{kJ/kmol.K}$.

Steps:

1. Calculate temperature change: $\Delta T = T_2 - T_1 = 333 \, \text{K} - 277 \, \text{K} = 56 \, \text{K}$

2. Change in internal energy ($\Delta U$): $\Delta U = n C_v \Delta T = (1 \, \text{kmol}) \times 21 \, \text{kJ/kmol.K} \times 56 \, \text{K}$ $\Delta U = 1176 \, \text{kJ}$

3. Change in enthalpy ($\Delta H$): $\Delta H = n C_p \Delta T = (1 \, \text{kmol}) \times 29.3 \, \text{kJ/kmol.K} \times 56 \, \text{K}$ $\Delta H = 1640.8 \, \text{kJ}$

Thus, the internal energy change is $1176 \, \text{kJ}$, and the enthalpy change is $1640.8 \, \text{kJ}$.

Would you like more details on any part of the solution? Here are a few related questions to consider:

1. How would the calculation change if the specific heats were temperature-dependent?
2. What assumptions are made about the ideal gas behavior in this calculation?
3. How does the change in pressure affect the internal energy and enthalpy?
4. What would happen if the process involved heat exchange with the surroundings?
5. How does the molar volume relate to the ideal gas law in this problem?

Tip: For ideal gases, internal energy depends only on temperature, not pressure.