Question

10 litres of a monoatomic ideal gas at 0⁰C and 10 atm pressure is suddenly released to 1 atm pressure and the gas expands adiabatically against this constant pressure. The final temperature and volume of the gas respectively are.

• A. T = 174.9 K, V = 64 L
• B. T = 153 K, V = 57 L
• C. T = 165.4 K, V = 78.8 L
• D. T = 161.2 K, V = 68.3 L

Answer 1

Answer: (A)

T = 174.9 K, V = 64 L

Answer 2

This is adiabatic irreversible process so, for this process

PV g = Constant, is not applicable

W = – Pext (V2 – V1)

But for adiabatic process

W = dU = $\left( \frac{P_{2}V_{2} - P_{1}V_{1}}{\gamma - 1} \right)$

PV = nRT

$\Rightarrow$ 10 × 10 = n × 0.082 × 273

$\Rightarrow$ n = 4.47 moles – Pext (V2 – V1) =$\left( \frac{P_{2}V_{2} - P_{1}V_{1}}{\gamma - 1} \right)$

$\Rightarrow$ – 1 × (V2 – 10) = $\frac{1 \times V_{2} - 10 \times 10}{1.67 - 1}$

$\Rightarrow$ (10 – V2) = $\frac{V_{2} - 100}{0.67}$

$\Rightarrow$ 6.7 – 0.67 V2 = V2 – 100

$\Rightarrow$106.7 = 1.67 V2

$\Rightarrow$ V2 = 64

$\therefore$nR [T2 – T1] = P2V2 – P1V1

$\Rightarrow$ 4.47 ´ 0.082 [T2 – 273] = 64 – 100 = – 36

$\Rightarrow$ (T2 – 273) = – 98.2

$\Rightarrow$ T2 = 174.8 K