An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $$V1$$ and contains ideal gas at pressure $$P1$$ and temperature $$T1$$ . The other chamber has volume $$V2$$ and contains ideal gas at pressure $$P2$$ and temperature $$T2$$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

Question

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $$V1$$  and contains ideal gas at pressure $$P1$$  and temperature $$T1$$ . The other chamber has volume  $$V2$$ and contains ideal gas at pressure  $$P2$$ and temperature  $$T2$$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

Infinity Learn · Accepted Answer

As no work is done and system is thermally insulated from surrounding it means sum of internal energy of gas in two partitions is constant , i.e.  $$U=U1+U2$$ Assuming both gases have same dof, then $$U=f(n1+n2)RT2And$$ $$U1=fn1RT12$$,$$U2=fn2RT22$$​Solving we get $$T=P1V1+P2V2T1T2P1V1T2+P2V2T1$$

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