Question

An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is :
(Take $C_v$ = 1.5 R, where R is gas constant )

• A. 0.32
• B. 0.24
• C. 0.15
• D. 0.08

Answer 1

Answer: (C)

0.15

$\text{Solution}:$

As we know the formula of work done which is equal to the

$W=P_0{V}_0$

Heat given will be equal to, = $Q_{AB}+Q_{BC}$

Now by using the heat given formula mentioned above, we can write it as

$\Rightarrow n{c}_{v}d{t}_{AB}+n{c}_{p}d{t}_{BC}$

As we already know the value of $C_v=1.5R$

Therefore substituting this value in the heat given equation, we get,

$=\frac{3}{2}\left(nR{T}_B-nR{T}_A\right)+\frac{5}{2}\left(nR{T}_C-nR{T}_B\right)$

For the monatomic gas, the value will be 3

Now putting the values we had calculated above, we get

$\Rightarrow \frac{3}{2}\left(2P_0{V}_0-P_0{V}_0\right)+\frac{5}{2}\left(4P_0{V}_0-2P_0{V}_0\right)$

On simplifying the solution, we get

$\Rightarrow \frac{13}{2}{P}_0{V}_0$

As we know the formula for efficiency can be given by

$\eta =\frac{W}{heat\text{gain}}$

It can also be written in the following way and also substituting the values, we get

$\Rightarrow \frac{P_0{V}_0}{\frac{\frac{13}{2}}{P_0{V}_0}}$

On solving,

$\Rightarrow \frac{2}{13}\Rightarrow \eta =0.15$

$\text{Hence C is the correct option}$