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Heat ;work and internal energy

Question

n moles of an ideal monoatomic gas undergoes a process from 14 as shown in the figure. T1=500K,  T4=200K,P1=105  Pa and P2=4×105Pa. In state 3, 3V3=V1. The change in internal energy from state 3 to state 4 is x Joules. The value of ‘x’ is (Take no. of moles n=1/R, where R = gas constant)

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Solution

P4×105T500=3×105300P=103T105nRTV=103T105

As  n=1R1V=103105T  straight line from 2 to 4At point 1 , V1=nRT1P1=500105m3 Given : 3V3=V1V3=V13=5003×105m3At point 3,1V3=103105T33×105500=1000105T3600=1000-105T3T3=250KChange in internal energy from state 3 to state 4 , U=nCVT4T3=32×200250 =75J



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