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An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $$V_1$$ and contains ideal gas at pressure $$P_1$$ and temperature $$T_1$$. 

The other chamber has volume $$V_2$$ and contain same ideal gas at pressure $$P_2$$ and temperature $$T_2$$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be:


A
T1T2(P1V1+P2V2)P1V1T2+P2V2T1
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B
P5V1T1+P2V2T2P1V1+P2V2
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C
P1V1T5+P2V2T1P1V1+P2V2
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D
T1T2(P1V1+P2V4)P1V1T1+P2V2T6
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Solution

The correct option is D $$\dfrac{T_1T_2(P_1V_1+ P_2V_2)}{P_1V_1T_2+ P_2V_2T_1}$$
Let the final temperature be $$T$$ and $$n_{1}$$ & $$n_{2}$$ be the no. of moles of gas in each container

Using conservation of energy

$$(n_{1}+_{2})C_{V}T=n_{1}C_{V}T_{1}+n_{2}C_{V}T_{2}$$

$$\Rightarrow\displaystyle \left(\frac{P_{1}V_{1}}{RT_{1}}+\frac{P_{2}V_{2}}{RT_{2}}\right)T=\frac{P_{1}V_{1}}{R}+\frac{P_{2}V_{2}}{R}$$

$$\Rightarrow \displaystyle \left(\frac{P_{1}V_{1}T_{2}+P_{2}V_{2}T_{1}}{T_{1}T_{2}}\right)T=P_{1}V_{1}+P_{2}V_{2}$$

$$\Rightarrow T=\dfrac{T_{1}T_{2}(P_{1}V_{1}+P_{2}V_{2})}{P_{1}V_{1}T_{2}+P_{2}V_{2}T_{1}}$$

Hence, the correct option is $$\text{A}$$

Chemistry

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