Question

To an ideal triatomic gas $800$ cal heat energy is given at constant pressure. If the vibrational mode is neglected, then energy used by gas in work done against surroundings is:

• A. $200$ cal
• B. $300$ cal
• C. $400$ cal
• D. $60$ cal

Answer 1

The correct option is C $400$ cal

Solution:- (C) $400cal$

According to the first law of thermodynamics,

$\Delta U=q+W$

$\Rightarrow W=\Delta U-q.....\left(1\right)$

Whereas $\Delta U$ is the change in the internal energy system and $W$ is the work done by the system.

As we know that,

$W=P\Delta V.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$P\Delta V=\Delta U-q$

$\Delta U=q+P\Delta V.....\left(3\right)$

As we know that,

$\Delta U=n{C}_v\Delta T$

Initial internal energy of $n$ moles of diatomic gas-

$U_1=n\left(\frac{6}{2}R\right)T_1=3nR{T}_1$

Internal energy of gas after heating-

$U_2=n\left(\frac{6}{2}\right)R{T}_2=3nR{T}_2$

$\therefore \Delta U=U_2-U_1$

$\Rightarrow \Delta U=3nR{T}_2-3NR{T}_1$

$\Rightarrow \Delta U=3nR(T_2-T_1)$

$\Rightarrow \Delta U=3nR\Delta T$

From ideal gas equation,

$PV=nRT$

$\Rightarrow P\Delta V=nR\Delta T$

$\therefore \Delta U=3P\Delta V.....\left(4\right)$

Now from ${eq}^n\left(3\right)\&\left(4\right)$, we have

$3P\Delta V=q+P\Delta V$

$2P\Delta V=q$

$\Rightarrow P\Delta V=\frac{q}{2}$

$\Rightarrow W=\frac{800}{2}=400cal\left(\text{From}\left(1\right)\right)$

Hence the work done is $400cal$.