JEE PYQ: Thermodynamics Question 24
Question 24 - 2020 (06 Sep Shift 1)
Initially a gas of diatomic molecules is contained in a cylinder of volume $V_1$ at a pressure $P_1$ and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume $2V_1$ is given by $P_2$. The ratio $P_2/P_1$ is __________.
Show Answer
Answer: 5
Solution
Using ideal gas equation, $PV = nRT$
$P_1 V_1 = nR \times 250$ $[\because T_1 = 250$ K] …(i)
$P_2(2V_1) = \frac{5n}{4}R \times 2000$ $[\because T_2 = 2000$ K] …(ii)
Dividing eq.(i) by (ii):
$\frac{P_1}{2P_2} = \frac{4 \times 250}{5 \times 2000} \Rightarrow \frac{P_1}{P_2} = \frac{1}{5}$
$\therefore \frac{P_2}{P_1} = 5$
BETA
