Question

$105$ calories of heat is required to raise the temperature of $3$ moles of an ideal gas at constant pressure from ${30}^{\circ }\text{C}$ to ${35}^{\circ }\text{C}$. The amount of heat required in calories to raise the temperature of the gas from ${60}^{\circ }\text{C}$ to ${65}^{\circ }\text{C}$ at constant volume is $[\gamma =\frac{C_P}{C_V}=1.4]$

• A. $50\text{cal}$
• B. $75\text{cal}$
• C. $70\text{cal}$
• D. $90\text{cal}$

Answer 1

The correct option is B $75\text{cal}$

At constant pressure, heat absorbed by the gas is given by
$\left(\Delta Q\right)=n{C}_P\Delta T=3\times C_P\times [35-30]=15C_P....\left(1\right)$
At constant volume, heat absorbed by the gas is given by $\left(\Delta U\right)=n{C}_V\Delta T=3\times C_V\times [65-60]=15C_V...\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$\frac{\Delta Q}{\Delta U}=\frac{C_P}{C_V}=\gamma$
From the data given in the question,
$\frac{\Delta Q}{\Delta U}=1.4$
$\Rightarrow \frac{105}{\Delta U}=1.4$
$\therefore \Delta U=75\text{cal}$
Thus, option (b) is the correct answer.