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Question: One mole of oxygen is contained in a rigid container with walls of inner surface area A, thickness d...

One mole of oxygen is contained in a rigid container with walls of inner surface area A, thickness d and thermal conductivity K. The gas is at initial temperature T0T_0 and the surrounding temperature of air is 2T. The temperature of the gas at time t is (R is gas constant) :-

A

T=T_0(2-e^{\frac{-2KA}{5dR}t})

B

T=T_0(2-e^{\frac{2KA}{5dR}t})

C

T=T_0e^{\frac{-2KA}{5dR}t}

D

T=T_0e^{\frac{KA}{2dR}t}

Answer

T=T_0(2-e^{\frac{-2KA}{5dR}t})

Explanation

Solution

The rate of heat transfer through the container wall is given by Fourier's Law: dQ/dt=−KA(T−2T0)/ddQ/dt = -KA(T - 2T_0)/d. The rate of change of internal energy of the gas is dQ/dt=nCVdT/dtdQ/dt = n C_V dT/dt. For one mole of oxygen (a diatomic gas), CV=52RC_V = \frac{5}{2}R. Equating these gives 52RdTdt=−KA(T−2T0)/d\frac{5}{2}R \frac{dT}{dt} = -KA(T - 2T_0)/d. This is a first-order linear differential equation. Solving it with the initial condition T(0)=T0T(0) = T_0 yields T(t)=2T0−T0e−2KA5Rdt=T0(2−e−2KA5Rdt)T(t) = 2T_0 - T_0 e^{-\frac{2KA}{5Rd}t} = T_0(2-e^{-\frac{2KA}{5Rd}t}).