Question

To an ideal non-linear triatomic gas, $800\text{cal}$ of heat energy is given at constant pressure. If vibrational modes are neglected, then the energy used by the gas in doing work against the surroundings is

• A. $200\text{cal}$
• B. $300\text{cal}$
• C. $400\text{cal}$
• D. $60\text{cal}$

Answer 1

The correct option is A $200\text{cal}$

Heat absorbed by a gas at constant pressure is given by
$\Delta Q=n{C}_P\Delta T.....\left(1\right)$
Work done during during isobaric process is
$\Delta W=P\Delta V=nR\Delta T....\left(2\right)$
Then, from $\left(1\right)$ and $\left(2\right)$
$\frac{\Delta W}{\Delta Q}=\frac{nR\Delta T}{n{C}_P\Delta T}$
$\Rightarrow \frac{\Delta W}{\Delta Q}=\frac{R}{C_P}$
We know that, $C_P=\frac{\gamma R}{\gamma -1}$
Thus,
$\frac{\Delta W}{\Delta Q}=\left(\frac{\gamma -1}{\gamma }\right)$
For an ideal non-linear triatomic molecule, $\gamma =\frac{4}{3}$.
And, from the data given in the question,
$\frac{\Delta W}{800}=1-\frac{1}{\left(\frac{4}{3}\right)}$
$\Rightarrow \frac{\Delta W}{800}=1-\frac{3}{4}$
$\Rightarrow \Delta W=200\text{cal}$
Thus, option (a) is the correct answer.