JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

One gas of $n_{1}$ mole of molecules at temperature $T_{1}$ , volume $V_{1}$ , and pressure $P_{1}$ , and another gas of $n_{2}$ mole of molecules at temperature $T_{2}$ , volume $V_{2}$ , and pressure $P_{2}$ , are mixed resulting in pressure $P$ and volume $V$ of the mixture. The temperature of the mixture is _________.

A
$(T_{1} + T_{2}) /2$
B
$T_{1} T_{2} P V / (T_{2} P_{1} V_{1} + T_{1} P_{2} V_{2})$
C
$(T_{2} P_{1} V_{1} + T_{1} P_{2} V_{2}) (T_{1} T_{2} P V)$
D
$∣ T_{1} - T_{2} ∣/2$

Solution

Apply $n = \frac{P V}{R T}$ to each gas and use mole conservation $n = n_{1} + n_{2}$ :

$\frac{P V}{R T} = \frac{P _{1} V _{1}}{R T _{1}} + \frac{P _{2} V _{2}}{R T _{2}} .$

Cancel $R$ and solve for $T$ :

$T = \frac{P V}{\frac{P _{1} V _{1}}{T _{1}} + \frac{P _{2} V _{2}}{T _{2}}} = \frac{T _{1} T _{2} P V}{T _{2} P _{1} V _{1} + T _{1} P _{2} V _{2}} .$

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