Heat And Thermodynamics Ques 122
- A cylinder with fixed capacity of $67.2 L$ contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by $20^{\circ} C$ is
[Take, $R=8.31 J mol^{-1} K^{-1}$ ]
(2019 Main, 10 April I)
(a) $700 J$
(b) $748 J$
(c) $374 J$
(d) $350 J$
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Answer:
Correct Answer: 122.(b)
Solution:
Formula:
- Given, capacity of cylinder is $=67.2 L$ and $\Delta T=20^{\circ} C$
At STP volume $=22.4 L$
$\therefore$ Number of moles $=\frac{67.2}{22.4}=3$
Now, change in heat is given as
$$ \Delta Q=n C _V \Delta T $$
Substituting the given values, we get
$$ \begin{aligned} \Delta Q & =3 \times \frac{3 R}{2} \times 20 \quad\left(\because \text { for He gas, } C _V=\frac{3}{2} R\right) \\ & =90 R J \end{aligned} $$
Given, $\quad R=8.31 J mol^{-1} K^{-1}$
$$ \therefore \quad \Delta Q=90 \times 8.31=747.9 J=748 J $$
BETA
