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Thermodynamics

Thermodynamics

by Kavitha G -
Number of replies: 2

One mole of monoatomic ideal gas at P=2 bar and T=273 K is compressed to 4 bar pressure following a reversible path obeying P/V=constant. Assume Cv = 12.5JK-1mol-1. The value of [(delta)u]/w for this process is minus.......

Answer : 3

Kindly explain.

In reply to Kavitha G

Re: Thermodynamics

by Rajat Rao -

P=kV. If P doubles, V also doubles. So, V changes from (say) V to 2V.

Integrate kV with respect to V within the limits V, 2V. This should give work done.

Work done = 1.5kV2 For getting the value of k, use the ideal gas equation PV=RT. Put P=kV and T=273(initial temperature). This gives kV2 = R(273) Substitute this value back in the expression for work done to get W = 1.5(R)(273). Now notice that 1.5R = Cv. So, work done W = Cv(273).

Use the ideal gas equation again to get the final temperature = 4(273). Change in internal energy = Cv(T2-T1). U = Cv(273)(4-1) = 3Cv(273).

Just divide U/W to get 3.
I don't see how the answer can be -3...