Question

A rigid tank contains $2 \, kg$ of an ideal gas at $500 \, kPa$ and $350 \, K$. A valve is opened and half of the gas is released. Then valve is closed. Final pressure = $300 \, kPa$. Find final temperature $T_2$ (in K).

Hint:

In rigid tanks with mass change, always use $\dfrac{P}{T} \propto m$.

Answer 1

Correct Answer: 420

Step 1: Apply ideal gas law initially.
\[ PV = mRT \Rightarrow \frac{P}{T} = \frac{mR}{V} = \text{constant (for rigid tank)}. \]
Step 2: Ratio form.
\[ \frac{P_1}{T_1} = \frac{m_1R}{V}, \frac{P_2}{T_2} = \frac{m_2R}{V} \] \[ \Rightarrow \frac{P_1}{T_1 m_1} = \frac{P_2}{T_2 m_2} \]
Step 3: Substitute values.
$P_1 = 500 \, kPa$, $T_1 = 350 \, K$, $m_1 = 2 \, kg$, $P_2 = 300 \, kPa$, $m_2 = 1 \, kg$. \[ \frac{500}{350 \times 2} = \frac{300}{T_2 \times 1} \] \[ T_2 = \frac{300 \times 350 \times 2}{500} = 420 \, K \] Correction: Since half mass is left, apply carefully: \[ T_2 = \frac{P_2 m_1 T_1}{P_1 m_2} \] \[ T_2 = \frac{300 \times 2 \times 350}{500 \times 1} = 420 \, K \] Final Answer: \[ \boxed{420 \, K} \]