Question

The initial internal energy of the gas in container 'A', if the containers were at room temperature 300K initially-

• A. 1406.25 cal
• B. 1000 cal
• C. 2812.5 cal
• D. none of these

Answer 1

Answer: (C)

2812.5 cal

Since the temperature rise and heat supplied to both the container is same.

$Q=n_A{C}_{VA}\Delta T=n_B{C}_{VB}\Delta T$ $\Rightarrow n_A{C}_{VA}=n_B{C}_{VB}-\left(1\right)$

also, $\Delta P_{A}V=n_{A}R\Delta T-\left(2\right)$

and, $\Delta P_{B}V=n_{B}R\Delta T-\left(3\right)$

Dividing $\left(2\right)$ by $\left(3\right)$,

$\frac{n_A}{n_B}=\frac{\Delta P_A}{\Delta P_B}=\frac{2.5}{1.5}=\frac{5}{3}$

$\Rightarrow n_A=\frac{5n_B}{3}$

From equation $\left(1\right)$

$\frac{C_{VB}}{C_{VA}}=\frac{5}{3}=\frac{\left(5R/2\right)}{\left(3R/2\right)}$

$C_{VB}=5R/2$, so $B$ is diatomic.

$C_{VA}=3R/2$, so $A$ is Mono-atomic.

Now,

$n_A=\frac{125}{40},C_{VA}=\frac{3R}{2}$

Internal energy is $U_i=n_A{C}_{VA}T$

$\Rightarrow U_i=\frac{125\times 3\times 2}{40\times 2}\times 300=2812.5cal$