A closed vessel contains $$0.1$$ mole of a monatomic ideal gas at $$200$$ K. If $$0.05$$ mole of the same gas at $$400$$ K is added to it, the final equilibrium temperature (in$$ $$K) of the gas in the vessel will be close to $$___________$$ .

Question

Infinity Learn · Accepted Answer

Assuming  $$Q=0$$,  $$W=0We$$ have      Δ $$U=0$$         ⇒   $$n1CV1Tf$$−$$T1+n2CV2Tf$$−$$T2=0$$⇒   $$n1CV1Tf$$−$$T1=-n2CV2Tf$$−$$T2=0$$⇒   $$n1Tf$$−$$T1=-n2Tf$$−$$T2$$∵For same $$gasCV1=CV2$$ ⇒   $$n1Tf-$$  $$n1T1=-n2Tf+n2T2$$⇒   $$n1Tf+n2Tf=$$  $$n1T1+n2T2$$⇒    $$Tf=n1T1+n2T2n1+n2$$        $$=0.1200+0.054000.1+0.05=20+200.15$$          $$=266.6666$$ K

Heat ;work and internal energy

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A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to ___________ .

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Heat ;work and internal energy

Remember concepts with our Masterclasses.

A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to ___________ .

Assuming Q=0, W=0We have Δ U=0 ⇒ n1CV1Tf−T1+n2CV2Tf−T2=0⇒ n1CV1Tf−T1=-n2CV2Tf−T2=0⇒ n1Tf−T1=-n2Tf−T2∵For same gasCV1=CV2 ⇒ n1Tf- n1T1=-n2Tf+n2T2⇒ n1Tf+n2Tf= n1T1+n2T2⇒ Tf=n1T1+n2T2n1+n2 =0.1200+0.054000.1+0.05=20+200.15 =266.6666 K

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