Question

An L.P.G cylinder weighs 14.8 kg when empty. It weighs 29 kg and shows a pressure of 2.5 atm. In the course used at $27{}^\circ C$ , the weight of the full cylinder was reduced to 23.3 kg. Find out the volume of n-butane in cubic meters used up to $27{}^\circ C$ and 1 atm. [Mol.mass of butane = 58].

Answer 1

The equation of state of a hypothetical ideal gas is given by the ideal gas law. The molecules of hypothetical ideal gas do not attract or repel each other. Ideal gas molecules take up no volume.

Complete Solution :
-The empirical relationships between the pressure, the volume, the temperature, and the number of moles of a gas is given by the ideal gas law. It can be used to calculate any of these properties if the other three are known.
-The equation for ideal gas law is as follows-
Ideal gas equation$= PV = nRT$
where R is the gas constant, i.e $R = 0.08206\dfrac{L.atm}{K.mol} = 8.3145\dfrac{J}{K.mol}$
P is the pressure
V is the volume
T is the temperature
N is the amount of substance.

-According to the question,
Weight of the cylinder with gas = 29 kg
Weight of the empty cylinder = 14.8 kg
Therefore, the weight of the gas in the cylinder $= 29 - 14.8 = 14.2kg$
Pressure in cylinder = 2.5 atm

-Now calculating the number of moles (n) in 14.2 kg $(14.2\times {{10}^{3}}g)$of butane,
We know that, $\text{no}\text{. of moles(n)}=\dfrac{\text{given mass}}{\text{molar mass}}=\dfrac{14.2\times {{10}^{3}}}{58}$
$\Rightarrow n=244.83mol$

-Applying gas equation for finding the volume of gas,
$V=\dfrac{nRT}{P}=\dfrac{2.4483\times 0.0821\times 300}{2.5}(\because 27{}^\circ C=273+27=300)$
$\Rightarrow V=2412\text{ liters}$

-Now, calculating the pressure in the cylinder after use:
Weight of cylinder after use = 23.2 kg
Weight of empty cylinder = 14.8 kg
Therefore the weight of the unused gas$=23.2 - 14.8 = 8.4kg(8.4\times {{10}^{3}}g)$

-From the gas equation, $P=\dfrac{nRT}{V}=\dfrac{\dfrac{8.4\times {{10}^{3}}}{58}\times 0.821\times 300}{2412}$
$\Rightarrow P = 1.478atm$
-Finally calculating the volume of gas used at 2.5 atm and $27{}^\circ C$ ,
Weight of the gas used $=14.2 - 8.4 = 5.8kg(5.8\times {{10}^{3}}g)$
From the gas equation, $V=\dfrac{nRT}{P}=\dfrac{5.8\times {{10}^{3}}\times 0.0821\times 300}{1}$
$\Rightarrow V = 246\text{ liters}=2.463{{m}^{3}}$

Note: The ideal gas law has some limitations. Since the ideal gas law assumes that gas particles have no volume and are not attracted to each other, if it is so, then on condensing an idea it should be able to condense to a zero volume. But in reality, the real gases do occupy some space, hence the gas law no longer applied here. Another limitation of gas law is that as the kinetic energy decreases while cooling the gas, the particles will eventually move slowly enough that their attractive forces cause them to condense.