Question

At a constant pressure, what should be the percentage increase in the temperature in Kelvin for a  $ 10\%  $  increase in the volume:

(A)  $ 10\%  $ 

(B)  $ 20\%  $ 

(C)  $ 5\%  $ 

(D)  $ 50\%  $ 

Answer 1

Hint: For solving this question, we need to use an equation involving the volume and temperature terms of a gas at initial and final steps at constant pressure. One equation that we are very familiar with is the equation proposed from Charles’s law. Charles’s law is a gas law, also called the law of volumes. It explains how gases tend to expand in volume when heated.

Complete answer:

Charles’s law states that:

“At constant pressure, the volume of a gas is directly proportional to its Kelvin temperature.”

 $  \Rightarrow  $   $ V \propto \,T $ 

Where V and T are the volume and temperature (taken in Kelvin) of the gas.

Hence,

 $  \Rightarrow V = kT $ 

Where k is the proportionality constant.

 $  \Rightarrow \dfrac{V}{T} = k, $  a constant

 $  \Rightarrow \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} $ 

Where  $ {{\text{V}}_{\text{1}}}{\text{ and }}{{\text{T}}_{\text{1}}} $  are the initial volume and temperature of the gas respectively, and  $ {{\text{V}}_2}{\text{ and }}{{\text{T}}_2} $  are the final volume and temperature respectively.

 In the question, we are asked to find the temperature increase in Kelvin when the volume of a gas increases by  $ 10\% . $ 

We are given:

 $  \Rightarrow {V_1}\, = {V_1} $ 

 $  \Rightarrow {T_1} = {T_1} $ 

 $  \Rightarrow {V_2}\, = 10\% \,\,{\text{greater than }}{V_1} $ 

 $  = {V_1} + 10\% \,\,{\text{of}}\,{V_1} $ 

 $  = {V_1} + \,\left( {\dfrac{{10}}{{100}} \times {V_1}} \right) = {V_1} + \,\left( {\dfrac{{{V_1}}}{{10}}} \right) $ 

 $  = \dfrac{{10{V_1} + {V_1}}}{{10}} $ 

 $  = \dfrac{{11}}{{10}}{V_1} $ 

 $  \Rightarrow {T_2} = ? $ 

 Substituting the values in Charles’s law,

 $  \Rightarrow \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} $ 

 $ \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{\left( {\dfrac{{11}}{{10}}{V_1}} \right)}}{{{T_2}}} $ 

 $ {T_2} = \left( {\dfrac{{\dfrac{{11}}{{10}}{V_1}}}{{{V_1}}}} \right) \times {T_1} $ 

 $  \Rightarrow {T_2} = \dfrac{{11}}{{10}}{T_1} $ 

 Since the final volume and final temperatures are  $ {V_2} = \dfrac{{11}}{{10}}{V_1} $  and  $ {T_2} = \dfrac{{11}}{{10}}{T_1} $  respectively and the percentage increase in volume is  $ 10\% , $  the temperature increase is also  $ 10\% . $ 

Therefore, the required answer is (A)  $ 10\% . $

Note:

Some other important gas laws are: Boyle’s law (describes relation between volume and pressure of a gas at constant temperature), Avogadro’s law (describes the relation between volume and number of moles of a gas at constant temperature and pressure), Gay Lussac’s law (describes relation between pressure and temperature of a gas at constant volume and mass).