Question

An LPG (liquified petroleum gas) cylinder weighs 14.8 kg when empty. When full, it weighs 29.0 kg and shows a pressure of 2.5 atm. In the course of use at 27°C, the weight of the full cylinder is reduced to 23.2 kg. Find out the volume of the gas in cubic metres used up at the normal usage conditions and the final pressure inside the cylinder. Assume LPG to be n-butane with normal boiling point of 0°C.

Answer 1

Calculating the volume of gas:

Given,

i) Weight of cylinder with gas = 29.0 kg

ii) Weight of empty cylinder = 14.8 kg

Therefore, the weight of gas in the cylinder = 29.0 - 14.8 = 14.2 kg

Pressure in cylinder = 2.5 atm

Therefore, number of moles (n) in 14.2 kg (14.2 × ${ 10 }^{ 3 }$ g) of butane

We know that, $n\quad =\quad \frac { Weight\quad of\quad butane }{ Molecular\quad weight\quad of\quad butane }=\quad \frac { 14.2\quad \times \quad { 10 }^{ 3 } }{ 58 }$  

n = 244.83 mol

Applying gas equation,

$V\quad=\quad \frac { nRT }{ P }$  

$=\quad \frac {244.83\quad \times \quad 0.0821\quad \times \quad 300 }{ 2.5 }$  

V = 2412 liters

[27°C  = 273 + 27 = 300]

Calculating the pressure in cylinder after use:

i) Weight of cylinder after use = 23.2 kg

ii) Weight of empty cylinder = 14.8 kg

Therefore weight of unused gas = 8.4 kg = 8.4 x ${ 10 }^{ 3 }$ g) moles of butane

We know that, $P\quad =\quad \frac { nRT }{ V }$

$=\frac {\frac {8.4\quad \times \quad { 10 }^{ 3 }}{ 58 }\quad \times \quad 0.0821\quad \times \quad 300}{2412}$

P = 1.478 atm

[V = 2412 L]

Calculating the volume of used gas at 2.5 atm and 27°C

Weight of the used gas = 14.2 - 8.4 = 5.8 kg

Pressure under normal usage conditions = 1 atm

$V\quad =\quad \frac { nRT }{ P }$

$=5.8\quad \times \quad \frac { 103 }{ 58 } \quad \times \quad 0.0821\quad \times \quad \frac { 300 }{ 1 } \quad \left[ \because \quad n\quad =\quad \frac { 5.8 }{ 58 }  \right]$

= 246 liters

V = 2.463 ${ m }^{ 3 }$