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← JEE Main 2026 · 4 Apr · Shift 1
PhysicsSingle correct+4 / -1Question 35 of 75

JEE Main 2026 · 4 Apr · Shift 1, Physics question 35

One gas of n1n_1 mole of molecules at temperature T1T_1, volume V1V_1, and pressure P1P_1, and another gas of n2n_2 mole of molecules at temperature T2T_2, volume V2V_2, and pressure P2P_2, are mixed resulting in pressure PP and volume VV of the mixture. The temperature of the mixture is ________.

Choose one option
Answer and worked solution

Official answer: B

Moles are conserved. n1+n2=P1V1RT1+P2V2RT2=PVRTn_1+n_2=\dfrac{P_1V_1}{RT_1}+\dfrac{P_2V_2}{RT_2}=\dfrac{PV}{RT}.

Solve for T.
T=PVP1V1T1+P2V2T2=T1T2PVT2P1V1+T1P2V2.T=\frac{PV}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}=\frac{T_1T_2PV}{T_2P_1V_1+T_1P_2V_2}.

(B)