Question

One mole of an ideal gas undergoes a process in which \[T = {T_0} + a{V^3}\], where \[{T_0}\] and ‘\[a\]’ are positive constants and \[V\] is volume. The volume for which the pressure of gas will be minimum is
A. \[{\left( {\dfrac{{{T_0}}}{{2a}}} \right)^{1/3}}\]
B. \[{\left( {\dfrac{{{T_0}}}{{3a}}} \right)^{1/3}}\]
C. \[{\left( {\dfrac{a}{{2{T_0}}}} \right)^{2/3}}\]
D. \[{\left( {\dfrac{a}{{3{T_0}}}} \right)^{2/3}}\]

Answer 1

Hint: Use the formula for ideal gas equation. This equation gives the relation between pressure, volume, temperature, number of moles and gas constant. Substitute the given value for the temperature and number of moles of the gas. Use the condition for having minimum pressure of the gas at constant volume of the gas.

Formula used:
The ideal gas equation is
\[PV = nRT\] …… (1)
Here, \[P\] is pressure of gas, \[V\] is volume of gas, \[n\] is the number of moles of gas, \[R\] is the gas constant and \[T\] is temperature of the gas.

Complete step by step answer:
We have given that the temperature of the ideal gas is given by
\[T = {T_0} + a{V^3}\]

Here, \[{T_0}\] and ‘\[a\]’ are positive constants and \[V\] is volume.

The number of moles of the gas is one.
\[n = 1\,{\text{mol}}\]

Substitute \[{T_0} + a{V^3}\] for \[T\] and \[1\,{\text{mol}}\] for \[n\] in equation (1).
\[PV = \left( {1\,{\text{mol}}} \right)R\left( {{T_0} + a{V^3}} \right)\]
\[ \Rightarrow PV = R{T_0} + aR{V^3}\]

Rearrange the above equation for pressure \[P\].
\[ \Rightarrow P = \dfrac{{R{T_0} + aR{V^3}}}{V}\]
\[ \Rightarrow P = \dfrac{{R{T_0}}}{V} + aR{V^2}\]

We have asked to determine the volume of the gas for which the pressure is minimum.

The pressure of the gas will be minimum if
\[\dfrac{{dP}}{{dV}} = 0\]

Substitute \[\dfrac{{R{T_0}}}{V} + aR{V^2}\] for \[P\] in the above equation.
\[\dfrac{{d\left[ {\dfrac{{R{T_0}}}{V} + aR{V^2}} \right]}}{{dV}} = 0\]
\[ \Rightarrow \dfrac{{d\left[ {\dfrac{{R{T_0}}}{V}} \right]}}{{dV}} + \dfrac{{d\left[ {aR{V^2}} \right]}}{{dV}} = 0\]
\[ \Rightarrow R{T_0}\dfrac{{d\left[ {\dfrac{1}{V}} \right]}}{{dV}} + aR\dfrac{{d{V^2}}}{{dV}} = 0\]
\[ \Rightarrow R{T_0}\left( {\dfrac{{ - 1}}{{{V^2}}}} \right) + aR\left( {2V} \right) = 0\]
\[ \Rightarrow \dfrac{{ - R{T_0}}}{{{V^2}}} + 2aRV = 0\]
\[ \Rightarrow 2a{V^3} = {T_0}\]

Rearrange the above equation for the volume of the gas.
\[ \Rightarrow V = {\left( {\dfrac{{{T_0}}}{{2a}}} \right)^{1/3}}\]

Therefore, the volume of the gas for which the pressure is minimum is \[{\left( {\dfrac{{{T_0}}}{{2a}}} \right)^{1/3}}\].

So, the correct answer is “Option A”.

Note:
The students should not get confused between the condition for minimum pressure of the gas at constant volume and the minimum volume of the gas at constant pressure. If this condition is not used properly, the final value for the volume of the gas will not be correct. Also, the students should be careful while taking derivatives.