A diatomic ideal gas expands from a volume of VA-1.00 mºto V, - 3.00 m along...

Question

Question

Answer 1

Answer #1

a] work done = area under curve = 0.5*(4e5 +2e5)*(2-1)+0.5*(4e5 +2e5)*(3-2) = 600000 J

b] Here initial and final Pressure is same, but V becomes 3 times, so T will become 3 times.

T2 = 3*T1

change in temperature = T2-T1 = 3T1-T1 = 2T1 = 2*P1V1/nR = 2*2e5*1/(67.3*8.314) = 715 K

c] change in internal energy = 5/2 nR deltaT = 5/2*67.3*8.314*715 = 1.0*10^6 J

d] total heat given = Work + internal energy change = 1.6*10^6 J

Answer 2

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