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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
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507.6k+ views
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.