JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Heat & Thermodynamics

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 ______.

Answer: (B)

Solution

By the ideal gas law the mole numbers are

n_{1} = \frac{P _{1} V _{1}}{R T _{1}}

and

n_{2} = \frac{P _{2} V _{2}}{R T _{2}}

. The mixture contains

n_{1} + n_{2}

moles, so

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

. Solving for the mixture temperature,

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

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option