Question

Find the amount of work done to increase the temperature of one mole of an ideal gas by .$30^\circ C$, if it is expanding under the condition $V \propto {T^{{2 \mathord{\left/ {{2 3}} \right.} 3}}}$. $\left( {R = 8.31J/mole/K} \right)$
(A) $16.62J$
(B) $166.2J$
(C) $1662J$
(D) $1.662J$

Answer 1

Hint
We need to find the pressure in terms of the volume from the ideal gas equation. Then after substituting the pressure in the equation of work done given as, $W = \int {PdV}$. Since the pressure is given as dependent on the temperature, we can find the ratio of $\dfrac{{dV}}{V}$ in terms of temperature and substitute in the equation for work done. On doing the integration we get the amount of work done.

Formula Used: In this solution we will be using the following formula,
$PV = nRT$
where $P$ is pressure and $V$is the volume.
$n$ is the number of moles, $R$ is the universal gas constant and $T$is the temperature.
$W = \int {PdV}$
where $W$ is the work done

Complete step by step answer
In the question we are given to be working with one mole of an ideal gas. So $n$ will be 1. Hence from the ideal gas equation we have,
$PV = RT$
Now keeping only the $P$ in the LHS we have,
$P = \dfrac{{RT}}{V}$
The work done by an ideal gas is given by the formula,
$W = \int {PdV}$
Here substituting the pressure we have,
$W = \int {\dfrac{{RT}}{V}dV}$……(1)
Now in the question we are said that the volume and the temperature are related as, $V \propto {T^{{2 \mathord{\left/ {{2 3}} \right.} 3}}}$
We can remove the constant of proportionality and get,
$V = C{T^{{2 \mathord{\left/ {{2 3}} \right.} 3}}}$ where $C$ is the constant of proportionality.
Now on taking derivative on both the sides of this equation we get,
$dV = d\left( {C{T^{{2 \mathord{\left/ {{2 3}} \right.} 3}}}} \right)$
So the RHS of this equation gives us,
$dV = C \times \dfrac{2}{3}{T^{\dfrac{2}{3} - 1}}dT$
This gives us,
$dV = C \times \dfrac{2}{3}{T^{\dfrac{{ - 1}}{3}}}dT$
Now we can take the ratio of $V$ and $dV$ and get,
$\dfrac{{dV}}{V} = \dfrac{{\dfrac{2}{3}C{T^{ - \dfrac{1}{3}}}dT}}{{C{T^{\dfrac{2}{3}}}}}$
On cancelling the $C$ from the numerator and the denominator we have,
$\dfrac{{dV}}{V} = \dfrac{2}{3}{T^{ - \dfrac{1}{3} - \dfrac{2}{3}}}dT$
So we get,
$\dfrac{{dV}}{V} = \dfrac{2}{3}{T^{ - 1}}dT$
We can substituting this value in equation 1 we get,
$W = \int {RT\dfrac{2}{3}{T^{ - 1}}dT}$
On cancelling the $T$,
$W = \dfrac{2}{3}R\int\limits_{{T_1}}^{{T_2}} {dT}$
We take the $\dfrac{2}{3}R$ out of the integration as it is a constant and set the limits over the integration from ${T_1}$ to ${T_2}$. On performing the integration we get,
$W = \dfrac{2}{3}R\left[ {{T_2} - {T_1}} \right]$
In the question we are given that the temperature increases by $30^\circ C$. So ${T_2} - {T_1} = 30^\circ C$. And it is given, $R = 8.31J/mole/K$. So substituting the values we get,
$W = \dfrac{2}{3} \times 8.31 \times 30$
On calculating this gives us,
$W = 166.2J$
So, the amount of work done to increase the temperature of one mole of an ideal gas by $30^\circ C$ is $166.2J$. Therefore the correct answer is option B.

Note
The ideal gas law or the gas equation is the equation of state of a hypothetical gas. Though it is a good approximation of various gasses under many conditions, still it has many limitations. In practice, most gasses like air are not ideal and follow the Joule-Thompson effect.