Q: A gas, for which γ is 4/3, is heated at constant pressure. The percentage of heat supplied used for external work is

(a) 25%

(b) 15%

(c) 60%

(d) 40%

Ans: (a)

Sol: P V = n R T

P ΔV = n R ΔT

ΔW = n R ΔT

ΔQ = ΔQp = n C_{p} ΔT

$\large \frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$

$\large = \frac{R}{C_p} = \frac{R}{\gamma R/(\gamma -1)}$

$\large = \frac{\gamma -1}{\gamma} = \frac{4/3 -1}{4/3}= \frac{1}{4}$

= 25 %

**Alternate : **

$\large \frac{\Delta U}{\Delta Q} = \frac{n C_v \Delta T}{n C_p \Delta T} $

$\large \frac{\Delta U}{\Delta Q} = \frac{1}{\gamma} = \frac{3}{4}$

$\large \frac{\Delta W}{\Delta Q} = \frac{\Delta Q – \Delta U}{\Delta Q} $

$\large \frac{\Delta W}{\Delta Q} = 1-\frac{\Delta U}{\Delta Q}= 1-\frac{3}{4} $

= 1/4 = 25%