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 Cp 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%