One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate

(a) the work done by the gas.
(b) the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB,
(c) the net heat absorbed by the gas in the path BC,
(d) the maximum temperature attained by the gas during the cycle.

Answer

A, B, C, D

Explanation

(a) `W=+` (Area of the cycle)
`=+1/2(2V_0-V_0)(3p_0-p_0)`
`=p_0V_0`
(b) `Q_(CA)=nC_pDeltaT`
`=n(5/2R)(T_A-T_C)`
`=5/2(p_AV_A-p_CV_C)`
`=5/2(p_0V_0-2p_0V_0)=-5/2p_0V_0`
`Q_(AB)=nC_VDeltaT`
`=n(3/2R)(T_B-T_A)`
`=3/2(p_BV_B-p_AV_A)`
`=3p_0V_0`
(c) `W_(BC)=+` (Area under the graph)
`=2p_0V_0`
`Delta_(BC)=nC_VDeltaT`
`=n(3/2R)(T_C-T_B)`
`=3/2(p_CV_C-p_BV_B)=-3/2p_0V_0`
`:.` `Q_(BC)=W_(BC)+DeltaU_(BC)=(p_0V_0)/(2)`
(d) p-V equation along path BC is
`p=-((2p_0)/(V_0))V+5p_0`
or `pV=-((2p_0)/(V_0))V^2+5p_0V`
or `RT=-((2p_0)/(V_0))V^2+5p_0V`
`:.` `T=1/R[5p_0V-(2p_0)/(V_0)V^2]`
For T to be maximum,
`(dT)/(dV)=0`
`:.` `5p_0-(4p_0)/(V_0)V=0`
`:.` `V=5/4V_0`
So at this volume, temperature is maximum.
Substituting this value of V in Eq. (i), we get
`T_(max)=(25p_0V_0)/(8R)`