Question

One mole of an ideal monatomic gas is taken round the cyclic process abca

Answer 1

Answer:

Tmax=25p0V0/8R

Explanation:

In order to understand the question, first see the figure in the attachment , where we have to find the maximum T on the entire path, which will be along the path BC.

p=−(2p0/V0)xV+5p0

or pV=−(2p0/V0)V2+5p0V

or RT=−(2p0/V0)V2+5p0V

∴∴ T=1/R[5p0V−2p0/V0 x V2] …………….eq (i)

For T to be maximum,

dT/dV=0

∴∴ 5p0−4p0/V0 x V = 0

∴∴ V=5/4V0

So at this volume, temperature is maximum.

Substituting this value of V in Eq. (i), we get

Tmax=25p0V0/8R