Question

A flask is filled with 13 gm of an ideal gas at 27⁰C and its temperature is raised to 52⁰C. The mass of the gas that has to be released to maintain the temperature of the gas in the flask at 52⁰C and the pressure remaining the same is

• A. 2.5 g
• B. 2.0 g
• C. 1.5 g
• D. 1.0 g

Answer 1

Answer: (D)

1.0 g

Answer 2

PV ∝ Mass of gas × Temperature

In this problem pressure and volume remains constant so $M_{1}T_{1}$= M₂T₂ = constant

∴$\frac{M_{2}}{M_{1}} = \frac{T_{1}}{T_{2}} = \frac{(27 + 273)}{(52 + 273)} = \frac{300}{325} = \frac{12}{13}$

⇒ $M_{2} = M_{1} \times \frac{12}{13} = 13 \times \frac{12}{13}gm = 12gm$

i.e. the mass of gas released from the flask = 13 gm – 12 gm = 1 gm.