Question

A gas occupies 3 litres at $0^\circ C$.What volume will it occupy at $- 20^\circ C$, pressure remaining constant?

Answer 1

We can make use of the Charle’s Law to solve this problem, which gives us an accurate relationship between the volume occupied by a gas and the absolute temperature, when the pressure and amount of gas is kept constant.
Formulas used: $\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$
Where $V$is the volume occupied by the gas, $T$ is the absolute temperature and the subscripts 1 and 2 denote the initial and final stages of the gas respectively.

Complete step by step solution:
As the pressure is kept constant in this particular question, and the same amount of gas is under consideration, we can make use of the Charle’s Law, which states that the volume occupied by a gas is directly proportional to the absolute temperature when pressure is kept constant. That is,
$V \propto {[T]_{P,n}}$
Where $V$ is the volume occupied by the gas, $T$ is the absolute temperature, $P$ is the pressure and $n$ is the amount of gas taken.
Therefore, the term ${V/T}$ will be a constant and considering an initial and final stage, we have the following result:
$\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} \Rightarrow {V_2} = \dfrac{{{V_1}{T_2}}}{{{T_1}}}$ ………………………..(1)
Where $V$ is the volume occupied by the gas, $T$ is the absolute temperature and the subscripts 1 and 2 denote the initial and final stages of the gas respectively.
First let us convert the temperatures into Kelvin by adding 273 to the Celsius temperatures.
$\Rightarrow 0^\circ C = 0 + 273 = 273K$ and $- 20^\circ C = - 20 + 273 = 253K$
Therefore, from the data we’ve been given,
${V_1} = 3L,{T_1} = 273K$ and ${T_2} = 253K$
Substituting these values into our equation (1), we get:
${V_2} = \dfrac{{3 \times 253}}{{273}}L$
On solving this, we get:
${V_2} = 2.78L$

Hence, at $- 20^\circ C$, the gas will occupy a volume of $2.78L$.

Note:
The temperature should be in absolute scale, that is, the unit should be Kelvin. If we fail to recognize the need to convert the unit, we may end up with absurd answers such as infinity.
Also note that the Charle’s Law is not applicable if the number of moles of the gas were to change during the process, even if pressure is kept constant. In this case, we will have to use the Ideal gas Law.