Question

An insulated rigid tank of volume 10 m$^3$ contains air initially at 1 MPa and 600 K. A valve connected to the tank is opened, and air is allowed to escape until the temperature inside the tank drops to 400 K. The temperature of the discharged air can be approximated as the average of the initial and final temperatures of the air in the tank. Neglect kinetic and potential energies of the discharged air. Assume that air behaves as an ideal gas with constant specific heat so that internal energy $u = c_p T$ and enthalpy $h = c_p T$. Then, the final pressure of the air in the tank is _________ MPa (round off to 2 decimal places). Assume: $c_p = 1.005$ kJ/(kg K), $\gamma = \frac{c_p}{c_v} = 1.4$

Hint:

For isochoric processes with ideal gases, use the temperature ratio to find the pressure ratio: $P_2/P_1 = T_2/T_1$.

Answer 1

Correct Answer: 0.2

The initial state of the air: \[ P_1 = 1\ \text{MPa}, \qquad T_1 = 600\ \text{K}, \qquad V = 10\ \text{m}^3 \] For an ideal gas, we use the relation: \[ PV = mRT \] We know the final temperature is $T_2 = 400$ K, and the process is isochoric (constant volume). Therefore, the pressure ratio can be derived as: \[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \] \[ P_2 = P_1 \times \frac{T_2}{T_1} = 1 \times \frac{400}{600} = 0.6667\ \text{MPa} \] Hence, the final pressure of the air in the tank is: \[ \boxed{0.67\ \text{MPa}} \]