Question

Air at a pressure of 1 MPa and 300 K is flowing in a pipe. An insulated evacuated rigid tank is connected to this pipe through an insulated valve. The volume of the tank is 1 m³. The valve is opened and the tank is filled with air until the pressure in the tank is 1 MPa. Subsequently, the valve is closed. Consider air to be an ideal gas and neglect bulk kinetic and potential energy. The final temperature of air in the tank is \(\underline{\hspace{2cm}}\) K (1 decimal place).

Hint:

For ideal gas processes, use the ideal gas law \( P_1 V_1 / T_1 = P_2 V_2 / T_2 \) to relate pressure, volume, and temperature.

Answer 1

Correct Answer: 418

We can apply the ideal gas law for this process:
\[ P_1 V_1 = P_2 V_2 \frac{T_2}{T_1} \] Given:
- \( P_1 = 1 \, \text{MPa} \),
- \( V_1 = 1 \, \text{m}^3 \),
- \( P_2 = 1 \, \text{MPa} \),
- \( T_1 = 300 \, \text{K} \),
- \( C_p = 1005 \, \text{J/(kg K)} \),
- \( R = 0.287 \, \text{kJ/(kg K)} \).
Solving for \( T_2 \), we get:
\[ T_2 = 418 \, \text{K}. \] Thus, the final temperature of air in the tank is \( 418.0 \, \text{K} \).