Question

An LPG cylinder contained 15 kg of
butane gas at 27 °C and 10 bar pressure.
The cylinder leaked and it was found
that after 4 days, the mass of the gas left
in the cylinder was only 10 kg. How much
drop in pressure has occurred in each
day?
0.52 bar
0.833 bar
1.6 bar
1.832 bar​

Answer 1

Given:

• Mass of butane gas = 15Kg = $15*10^{3}g$  
• Mass of butane gas after four days = 10Kg = $10*10^{3}g$
• Temperature = 27°C = 300K
• Pressure ($P_1$) = 10 bar

To Find:

• Pressure drop per day.

Solution:

• To find pressure we have to use the formula, PV = nRT
• To find that we need to find volume.
• we have the molecular weight of butane($C_4H_10$) = 48+10 = 58
• So, $n_1 = \frac{15*10^{3}}{58} = 258.6$ and $n_2 = \frac{10*10^{3} }{58}=172.4$
• We need to find the volume by using $P_1V_1=n_1RT$
• Substitute the values we get, $V_1 = \frac{n_1RT}{P_1} = \frac{258.6*0.0838*300}{10} = 643.91$
• We have $n_1$ and $n_2$ but we need n = $n_1-n_2 = 86.2$
• n = 86.2
• To find Pressure Drop (P), we are using $P = \frac{nRT}{V}$ = $\frac{86.2*0.083*300}{643.91} = \frac{2146.38}{643.91} = 3.333$ bar.
• Pressure drop per day = P/4 = 3.33/4 = 0.833 bar

The Pressure per day is 0.833 bar.