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 _________.
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answer is 25.50.
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Detailed Solution
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