Question

An LPG cylinder contains 15 kg butane at 27℃ and 10 atm pressure. It is leaking. If its pressure decreases to 8 atm after one day, then the quantity of gas leaked is:

• A. 2 kg
• B. 1 kg
• C. 4 kg
• D. 3 kg

Answer 1

The correct option is D

3 kg

Explanation:

Step-1: Calculate the amount of the gas in the cylinder after leaking

Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$

• Additionally, $\mathrm{n}=\frac{\mathrm{Given}\mathrm{mass}\left(\mathrm{m}\right)}{\mathrm{Molar}\mathrm{mass}\left(\mathrm{M}\right)}$
• Therefore by substituting, $\mathrm{PV}=\frac{\mathrm{mRT}}{\mathrm{M}}$

where P= Pressure of the gas, V= Volume of the gas, T= Temperature, R= Ideal gas constant

By equating the values before and after the leakage,

• $\frac{{\mathrm{P}}_1{\mathrm{V}}_1{\mathrm{M}}_1}{{\mathrm{m}}_1\mathrm{RT}}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2{\mathrm{M}}_2}{{\mathrm{m}}_2\mathrm{RT}}$
• Given, ${\mathrm{V}}_1={\mathrm{V}}_2=\mathrm{V}$as the volume of LPG cylinder is fixed
• ${\mathrm{M}}_1={\mathrm{M}}_2=\mathrm{M}$ since Molar mass of Butane will be same
• ${\mathrm{T}}_1={\mathrm{T}}_2=T$
  ${\mathrm{m}}_1=15\mathrm{kg},{\mathrm{m}}_2=?$
• $\frac{10\mathrm{VM}}{15\mathrm{RT}}=\frac{8\mathrm{VM}}{{\mathrm{m}}_2\mathrm{RT}}$
• $\frac{10}{15}=\frac{8}{{\mathrm{m}}_2}$
• ${\mathrm{m}}_2=\frac{8\times 15}{10}=12\mathrm{kg}$

Step-2: Calculate the amount of gas leaked

Amount of gas leaked= Amount of gas present in the cylinder initially- Amount of gas present in the cylinder after leakage

• Quantity of gas leaked= 15 -12 = 3 kg

Hence, option (D) is correct.