Question

An LPG cylinder, containing 15 kg butane at \[27^\circ {\text{C and 10 atm}}\] pressure, is leaking. After one day, its pressure decreases to 8 atm. The quantity of gas leaked is:
A. 1 kg
B. 2 kg
C. 3 kg
D. 4 kg

Answer 1

Hint: The ideal gas law gives the relationship between the pressure P, the volume V, the number of moles n, the temperature T and the ideal gas constant R. Write the expression for the ideal gas law.
PV = nRT.

Complete step by step answer:
The number of moles n are obtained when given mass w is divided with molecular weight M.
Substitute \[n = \dfrac{w}{M}\] in the ideal gas equation
\[PV = nRT \\
PV = \dfrac{w}{M}RT \\
\dfrac{P}{w} = \dfrac{{RT}}{{MV}} \\\]
At constant temperature and pressure, for a given gas, the term \[\dfrac{{RT}}{{MV}}\] is a constant
R is an ideal gas constant and M is constant if the gas remains the same.
Hence, the equation becomes
\[\dfrac{P}{w} = K\]
For initial and final conditions, the equation becomes
\[\dfrac{{{P_i}}}{{{w_i}}} = \dfrac{{{P_f}}}{{{w_f}}}\]
Here, the subscripts i and f represent initial and final values.
Rearrange the above equation.
\[{w_f} = {P_f} \times \dfrac{{{w_i}}}{{{P_i}}}\]
Substitute values in the above equation and calculate the final mass of gas present in the cylinder.
\[{w_f} = {P_f} \times \dfrac{{{w_i}}}{{{P_i}}} \\
= 8 \times \dfrac{{15}}{{10}} \\
= 12{\text{ kg}} \\ \]
Calculate the quantity of gas leaked by subtracting the final mass from the initial mass.
\[{w_i} - {w_f} = 15{\text{ kg}} - 12{\text{ kg}} \\
= 3{\text{ kg}} \\\]
In one day, 3 kg of butane gas is leaked.

Hence, the option C ) is the correct option.

Note: There is another way to solve the problem. The difference in initial and final pressures is 2 atm. It is 20 % of the initial value. Hence, the amount of gas leaked will also be 20 % of the initial value 15 kg which comes out to be 3 kg.