Question

Consider 1 kg of an ideal gas at 1 bar and 300 K contained in a rigid and perfectly insulated container. The specific heat at constant volume is \( c_v = 750\ \text{J kg}^{-1}\text{K}^{-1} \). A stirrer performs 225 kJ of work on the gas. Assume no heat interaction. The final pressure of the gas will be ________________ bar (integer).

Hint:

In rigid adiabatic systems, all work done appears as an increase in internal energy, meaning temperature rises directly with \(W = m c_v \Delta T\).

Answer 1

Correct Answer: 2

Since the container is perfectly insulated and rigid, the process is: - Adiabatic (\(Q = 0\))
- Constant volume (\(V = \text{constant}\))
The first law for a closed system gives:
\[ \Delta U = Q + W = 0 + 225\ \text{kJ} = 225\ \text{kJ} \] For an ideal gas, the change in internal energy is:
\[ \Delta U = m c_v \Delta T \] Given: \[ m = 1\ \text{kg}, \qquad c_v = 750\ \text{J/kg·K} = 0.75\ \text{kJ/kg·K} \] So, \[ 225 = 1 \times 0.75 \times \Delta T \] \[ \Delta T = \frac{225}{0.75} = 300\ \text{K} \] Initial temperature: \[ T_1 = 300\ \text{K} \] Final temperature: \[ T_2 = T_1 + \Delta T = 300 + 300 = 600\ \text{K} \] Since volume is constant, pressure is proportional to temperature:
\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \] \[ P_2 = 1\ \text{bar} \times \frac{600}{300} = 2\ \text{bar} \] Thus, the final pressure of the gas is \(2\) bar.