JEE PYQ: Thermodynamics Question 9
Question 9 - 2021 (25 Feb Shift 2)
A thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If $V_2 = 2V_1$, then the ratio of temperature $T_2/T_1$ is :
[P-V diagram showing: point 1$(P_1, V_1, T_1)$ connected to point 2$(P_2, V_2, T_2)$ by curve $PV^{1/2} =$ constant]
(1) $\frac{1}{\sqrt{2}}$
(2) $\frac{1}{2}$
(3) 2
(4) $\sqrt{2}$
Show Answer
Answer: (4)
Solution
From P-V diagram, given $PV^{1/2} =$ constant …(i)
We know that $PV = nRT$, so $P \propto \frac{T}{V}$
Put in equation (i): $\frac{T}{V}(V)^{1/2} =$ constant
$T \propto V^{1/2}$
$\frac{T_2}{T_1} = \left(\frac{V_2}{V_1}\right)^{1/2} = \sqrt{2}$
BETA
