Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in the P-T diagram. During the process AB , pressure and temperature of the gas very such that `PT=Constant`. It `T_1=300K`, calculate

(a) the work done on the gas in the process AB and
(b) the heat absorbed or released by the gas in each of the processes. Give answer in terms of the gas constant R.

Explanation

For the process A - B, it is given that
PT = constant
Differentiating above equation partially, we have
PdT + TdP = 0
Equation of state for two moles of a gas
`PV = 2 RT` or `P = (2RT)/(V)`
After differentiating Eq. (ii) partially, we get
PdV + VdP = 2R dT
From Eq. (ii) partially, we get
PdV + VdP = 2R dT
From Eqs. (i) and (ii), we have
`((2RT)/(V)) dT + T dP = 0`
or 2RT dT + VTdP = 0
VdP = - 2 RdT
Now form Eqs. (iii) and (iv), we have
`- 2RdT + VdP = 2RdT`
or `PdV = 4 RdT`
a. The work done in the process AB
`W_(AB) int PdV = int_(600)^(300) 4 RdT`
`= 4 R |T|_(600)^(300) = 4 R (300 - 600)`
`= - 1200 R`
b. As process `B rarr C` is isobaric, so
`Q_(BC) = nC_(P) Delta T = 2 xx (5R)/(2) xx (600 - 300)`
= 1500 R
ii. Process `C rarr A` is isothermal, so `Delta U = 0`
`Q_(CA) = Delta U + W_(CA) = W-(CA)`
`W_(CA) = nRT 1n (P_(C ) // P_(A))`
`2R xx 600 1n (2P_(1) // P_(1)) = 1200 R1n 2`
`Q_(CA) = 1200 R 1n 2`
Again for process `A rarr B`
`Q_(AB) = Delta U + W_(AB)`
`= nC_(V) Delta T + W_(AB)`
`= 2 xx ((3R)/(2)) xx (300 - 600) - 1200 R`
`= - 900 R - 1200 R = - 2100 R`