Question

1/2 mole of helium is contained in a container at STP. How much heat energy is needed to double the pressure of the gas, (volume is constant) heat capacity of gas is 3 J g–1 K–1.

• A. 1436 J
• B. 736 J
• C. 1638 J
• D. 5698 J

Answer 1

Answer: (C)

1638 J

Answer 2

Here, $n = \frac{1}{2},c_{v} = 3Jg^{- 1}K^{- 1},M = 4$

$\therefore C_{V} = Mc_{v} = 4 \times 3 = 12mol^{- 1}K^{- 1}$

At constant volume$P \propto T.$

$\therefore\frac{P_{2}}{P_{1}} = \frac{T_{2}}{T_{1}} = 2,T_{2} = 2T_{1}$

Rise in temperature $\Delta T = T_{2} - T_{1}$

$= 2T_{1} - T_{1} = T_{1} = 273K$

Heat required, $\Delta Q = nC_{V}\Delta T$

$= \frac{1}{2} \times 12 \times 273 = 1638J$