Two moles of helium gas undergo a cyclic process as shown in Fig. Assuming the gas to be ideal, calculate the following quantities in this process

(a) The net change in the heat energy
(b) The net work done
(c) The net change in internal energy

Answer

(i) `1152 J (ii) 1152J (iii) zero`

Explanation

`V_(A) = V_(0) = (nRT_(A))/(P_(A)) = (2R xx 300)/(2) = 300R`
`V_(B) = (nRT_(B))/(P_(B)) = (2R xx 400)/(2) = 400R = (4)/(3)V_(0)`
`V_(C) = (nRT_(C))/(P_(C)) = (2R xx 400)/(1) = 800R = (8)/(3)V_(0)`
`V_(D) = (nRT_(D))/(P_(D)) = (2R xx 300)/(1) = 600R = 2V_(0)`
For cyclic process `DeltaU = 0 , Q=W`
`W_(A rarr B) = P_(A) (V_(B) - V_(A)) = 2((4V_(0))/(3) - V_(0)) xx 10^(5) = (2V_(0))/(3) xx 10^(5)`
`W_(B rarrC) = P_(B) V_(B) ln2 = 2 xx 10^(5) xx (4V_(0))/(3)ln2 = (8V_(0))/(3)ln2 xx 10^(5)`
`W_(CrarrD) = P_(C)(V_(D) - V_(C)) = 10^(5) (2V_(0) - (8V_(0))/(3)) = (2V_(0))/(3)xx 10^(5)`
`W_(DrarrA) =-P_(D)V_(D).ln2 = -10^(5)xx 2 xxln2`
`therefore W = W_(ArarrB) + W_(BrarrC) + W_(CrarrD) + W_(DrarrA) = 1152J`
(i) `therefore Q= 1152J` (ii) `W=1152J` (iii) `DeltaU = 0`