A container is divided into two chambers by partition. The volume of first chamber is $$4.5$$ litre and second chamber is $$5.5$$ litre. The first chamber contain $$3.0$$ moles of gas at pressure $$2.0$$ atm and second chamber contain $$4.0$$ moles of gas at pressure $$3.0$$ atm. After the partition is removed and the mixture attains equilibrium pressure existing in the mixture is x×$$10$$−$$1atm$$. Value of x is $$_________$$.

Question

A container is divided into two chambers by partition. The volume of first chamber is $$4.5$$ litre and second chamber is $$5.5$$ litre. The first chamber contain $$3.0$$ moles of gas at pressure $$2.0$$ atm and second chamber contain $$4.0$$ moles of gas at pressure $$3.0$$ atm. After the partition is removed and the mixture attains equilibrium pressure existing in the mixture is x×$$10$$−$$1atm$$. Value of x is $$_________$$.

Infinity Learn · Accepted Answer

Using formula, $$U=nCVT=nfR2T=f2PV$$ Internal energy of system is conserved. ⇒$$f2(2$$×$$4.5)+f2(3$$×$$5.5)=f2P(5.5+4.5)$$⇒$$9+16.5=P(10)$$⇒$$P=2.55atm$$⇒$$P=25.5$$×$$10$$−$$1atm$$

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