Question

when an ideal diatomic gas is heated at constant pressure, its initial energy increased by $50cal$ then the work done by the gas is:

• A. $30cal$
• B. $50cal$
• C. $70cal$
• D. $20cal$

Answer 1

Answer: (D)

$20cal$

Given: Ideal diatomic gas, $△U$= 50 cal

Solution:

For ideal diatomic gas $C_p=\frac{7R}{2}$ and $C_v=\frac{5R}{2}$

So, $\gamma =\frac{7}{5}$

We know that, $△Q=n{C}_p△T$

Also $△U=n{C}_v△T$

So, $\frac{△Q}{△U}=\frac{n{C}_p△T}{n{C}_v△T}=\frac{C_p}{C_v}=\gamma$

$\therefore \frac{△Q}{50}=\frac{7}{5}$

$\Rightarrow △Q=70cal$

From first law of thermodynamics, $△Q=△U+W$

Putting the values of $△Q=70cal$ and $△U=50cal$, we get,

W=20 cal

Hence, the correct option is D.