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Question: At a constant pressure, what should be the percentage increase in the temperature in Kelvin for a \(...

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

(A) 10%10\%

(B) 20%20\%

(C) 5%5\%

(D) 50%50\%

Explanation

Solution

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 VTV \propto \,T

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

Hence,

V=kT\Rightarrow V = kT

Where k is the proportionality constant.

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

V1T1=V2T2\Rightarrow \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}

Where V1 and T1{{\text{V}}_{\text{1}}}{\text{ and }}{{\text{T}}_{\text{1}}} are the initial volume and temperature of the gas respectively, and V2 and T2{{\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%.10\% .

We are given:

V1=V1\Rightarrow {V_1}\, = {V_1}

T1=T1\Rightarrow {T_1} = {T_1}

V2=10%greater than V1\Rightarrow {V_2}\, = 10\% \,\,{\text{greater than }}{V_1}

=V1+10%ofV1= {V_1} + 10\% \,\,{\text{of}}\,{V_1}

=V1+(10100×V1)=V1+(V110)= {V_1} + \,\left( {\dfrac{{10}}{{100}} \times {V_1}} \right) = {V_1} + \,\left( {\dfrac{{{V_1}}}{{10}}} \right)

=10V1+V110= \dfrac{{10{V_1} + {V_1}}}{{10}}

=1110V1= \dfrac{{11}}{{10}}{V_1}

T2=?\Rightarrow {T_2} = ?

Substituting the values in Charles’s law,

V1T1=V2T2\Rightarrow \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}

V1T1=(1110V1)T2\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{\left( {\dfrac{{11}}{{10}}{V_1}} \right)}}{{{T_2}}}

T2=(1110V1V1)×T1{T_2} = \left( {\dfrac{{\dfrac{{11}}{{10}}{V_1}}}{{{V_1}}}} \right) \times {T_1}

T2=1110T1\Rightarrow {T_2} = \dfrac{{11}}{{10}}{T_1}

Since the final volume and final temperatures are V2=1110V1{V_2} = \dfrac{{11}}{{10}}{V_1} and T2=1110T1{T_2} = \dfrac{{11}}{{10}}{T_1} respectively and the percentage increase in volume is 10%,10\% , the temperature increase is also 10%.10\% .

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