Question

0.5 m3 of air at pressure of 0.3 bar is compressed isothermally to 0.1
m3.Calculate final pressure, heat supplied, work done and change in intemal
energy.
Solution of this please.​

Answer 1

Answer:

For isothermal process,

P1V1 = P2V2

0.5×0.3 = P2 × 0.1

Final pressure, P2 = 1.5 bar

In isothermal process there is no change in temperature hence no change in internal energy(U).

So, change in internal energy, dU = 0.

dQ = dU + dW

dQ = dW ( isothermal process, dU = 0)

dQ = P1V1 ln(P1/P2) = P1V1 ln(V2/V1)

dQ = 50000×0.3 × ln(1/3) = - 16479.1 j

dQ = -16.5 kj

So, final pressure , P2 = 1.5 bar

Heat supplied, dQ = -16.5 kj

Work done dW = dQ = -16.5 kj

Change in internal energy, dU = 0.

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