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Question: A gas occupies \(500c{m^3}\) at normal temperature. At what temperature will the volume of the gas b...

A gas occupies 500cm3500c{m^3} at normal temperature. At what temperature will the volume of the gas be reduced by 20%20\% of its original volume, pressure being constant?

Explanation

Solution

In the above given question we asked to find out the temperature (T2)({T_2}) and the volume (V1)({V_1}) and (V2)\left( {{V_2}} \right) and (T1)\left( {{T_1}} \right) is given. Implementing all these given data to the equation of Charles law we will get the desired value of (T2)({T_2}) .

Complete answer: Given,
The volume of an ideal gas is directly proportional to the absolute temperature at constant pressure, according to Charles law. When the pressure imposed on a sample of a dry gas is held constant, the Kelvin temperature and volume will be in direct proportion, according to the law.
Charles' Law, often known as the law of volumes, describes how a gas expands as the temperature rises. A reduction in temperature, on the other hand, will result in a drop in volume.
When comparing a substance under two different situations, we can write the following from the above statement:

OR  V2T1=T2T1  OR{\text{ }} \\\ \dfrac{{{V_2}}}{{{T_1}}} = \dfrac{{{T_2}}}{{{T_1}}} \\\

This equation shows that when the absolute temperature of the gas rises, the volume of the gas rises in proportion.
Initial volume of the gas V1{V_1} = 500cm3500c{m^3}
Normal temperature T1{T_1} = 273K273K
The Volume is reduced by 20%20\% of the initial volume, i.e volume which remains is (100−20)% = 80%\left( {100 - 20} \right)\% {\text{ }} = {\text{ }}80\%
Therefore,

=80100×500=400cm3 { = \dfrac{{80}}{{100}} \times 500 = 400c{m^3}}

Thus V2 =400cm3{V_2}{\text{ }} = 400c{m^3}
We have to find out T2{T_2}
From Charles Law we know that,
V1T1‘= V2T2{V_1}{T_{1`}} = {\text{ }}{V_2}{T_2}
Now, Substituting the values we get
500×273=400×T2500 \times 273 = 400 \times T_2
Thus,

  T2 = 218.4K {\;T_2{\text{ }} = {\text{ }}218.4K}

Therefore, at 218.4K218.4K temperature the volume of the gas is reduced by 20%20\% of its original volume.

Note:
The only variable that is modified is temperature, because pressure is kept constant. This means that we may compare volume and temperature using Charles' law. Volume and temperature are directly proportional since they are on opposite sides of the ideal gas law.