Question

Air at 400 K and 200 kPa is heated at constant pressure to 600 K. Assuming internal energy is a function of temperature only, the magnitude of change in internal energy during this process is __________ kJ/kmol (rounded off to one decimal place). Molar specific heat of air at constant volume: \[ \bar{c}_v = a + bT + cT^2 \] where $a = 19.686 \, \text{kJ/kmol-K}$, $b = 0.002 \, \text{kJ/kmol-K}^2$, $c = 0.5 \times 10^{-5} \, \text{kJ/kmol-K}^3$.

Hint:

For real gases with temperature-dependent specific heats, always integrate $c_v(T)$ rather than assuming constant $c_v$.

Answer 1

Correct Answer: 4389

Step 1: Internal energy change expression.
The change in internal energy is given by: \[ \Delta U = \int_{T_1}^{T_2} \bar{c}_v \, dT \]
Step 2: Substitute $\bar{c_v$.}
\[ \Delta U = \int_{400}^{600} \left( a + bT + cT^2 \right) dT \]
Step 3: Perform integration.
\[ \Delta U = a(T_2 - T_1) + \frac{b}{2}(T_2^2 - T_1^2) + \frac{c}{3}(T_2^3 - T_1^3) \]
Step 4: Substitution of values.
- $a = 19.686$, $b = 0.002$, $c = 0.5 \times 10^{-5}$
- $T_1 = 400$, $T_2 = 600$ \[ \Delta U = 19.686(600 - 400) + \frac{0.002}{2}(600^2 - 400^2) + \frac{0.5 \times 10^{-5}}{3}(600^3 - 400^3) \] \[ \Delta U = 19.686(200) + 0.001(360000 - 160000) + \frac{0.5 \times 10^{-5}}{3}(216 \times 10^6 - 64 \times 10^6) \] \[ \Delta U = 3937.2 + 200 + 86.67 = 4223.87 \, \text{kJ/kmol} \] Final Answer: \[ \boxed{4224.0 \, \text{kJ/kmol}} \]