Question

If a gas expands with temperature according to the relation $V=k{{T}^{\dfrac{2}{3}}}$, then what will be the work done, when the temperature changes by ${{30}^{o}}C$?
A. $40R$
B. $30R$
C. $20R$
D. $10R$

Answer 1

This question is related to adiabatic processes. So we are going to use the First Law of Thermodynamics and the ideal gas equation to solve this problem. Adiabatic processes, in thermodynamics, change occurring within a system as a result of transfer of energy to or from the system in the form of work only; i.e., no heat is transferred.

Complete answer:
As we know that for an adiabatic process, no heat transfer takes place, so the value of $dq$ for the system will be zero. Now, according to the first law of thermodynamics:
$dU=dq+dw$
$\Rightarrow dU=dw\ \ \ \ \ \ \left[ \because dq=0 \right]$
$\Rightarrow {{C}_{v}}dT=-pdV\ \ \ \ \ \ \left[ \because dU={{C}_{v}}dT\ \ \text{and }dw=-pdv \right]\ \ \ \ \ \ \ \ \ ...(1)$
As per ideal gas equation, $pV=nRT$
$\Rightarrow p=\dfrac{nRT}{V}$
Substituting value of pressure in equation (1):
${{C}_{v}}dT=-\dfrac{nRT}{V}dV$
$\Rightarrow \dfrac{{{C}_{v}}dT}{T}=-nR\dfrac{dV}{V}$
Integrating both sides by applying proper limits, we get the relation between volume and temperature as follows:
$\Rightarrow {{C}_{v}}\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{\dfrac{dT}{T}=-nR\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{\dfrac{dV}{V}}}$
$\Rightarrow {{C}_{v}}\left[ \ln T \right]_{{{T}_{1}}}^{{{T}_{2}}}=-nR\left[ \ln V \right]_{{{V}_{1}}}^{{{V}_{2}}}$
$\Rightarrow {{C}_{v}}\ln \left( \dfrac{{{T}_{2}}}{{{T}_{1}}} \right)=-nR\ln \left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)$
Applying properties of logarithmic function:
$\Rightarrow \ln {{\left( \dfrac{{{T}_{2}}}{{{T}_{1}}} \right)}^{{{C}_{v}}}}=\ln {{\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)}^{-nR}}\ \ \ \ \ \ \ \ \left[ \because b\ln a=\ln {{a}^{b}} \right]$
Taking antilog, the expression will be as follows:
$\Rightarrow {{\left( \dfrac{{{T}_{2}}}{{{T}_{1}}} \right)}^{{{C}_{v}}}}={{\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)}^{-nR}}$
For one mole of gas, $n=1$ and $R={{C}_{p}}-{{C}_{v}}$. Substituting the value of n and R in the expression:
$\Rightarrow {{\left( \dfrac{{{T}_{2}}}{{{T}_{1}}} \right)}^{{{C}_{v}}}}={{\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)}^{{{C}_{v}}-{{C}_{p}}}}$
$\Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}={{\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)}^{1-\dfrac{{{C}_{p}}}{{{C}_{v}}}}}$
$\Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}={{\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)}^{1-\gamma }}\ \ \ \ \ \left[ \because \gamma =\dfrac{{{C}_{p}}}{{{C}_{v}}} \right]$
In general, this equation can be represented as $T{{V}^{\gamma -1}}=\text{constant}$ and it is known as Poisson’s equation. Alternatively, the expression can also be represented as follows:
${{T}^{\dfrac{1}{1-\gamma }}}V=k\ \ \ \ \ ...(2)$
Where k is any constant.
The relation of volume and temperature given in the question corresponds to Poisson’s equation and thus, we can say that the system here represents reversible adiabatic process.
Now, as per question the relation between volume and temperature is given as follows:
$V=k{{T}^{\dfrac{2}{3}}}$
$\Rightarrow \dfrac{V}{{{T}^{\dfrac{2}{3}}}}=k$
$\Rightarrow V{{T}^{-\dfrac{2}{3}}}=k$
Comparing the above expression with equation (2), the value of $\gamma$ will be as follows:
$\dfrac{1}{1-\gamma }=-\dfrac{2}{3}$
$\Rightarrow 2\gamma -2=3$
$\Rightarrow \gamma =\dfrac{5}{2}$
Work done in a reversible adiabatic process is calculated using a formula which is given as follows:
$W=\dfrac{R({{T}_{2}}-{{T}_{1}})}{\gamma -1}$
As per question, the change in temperature i.e., ${{T}_{2}}-{{T}_{1}}={{30}^{o}}C$. Substituting the value of temperature and $\gamma$, the work done for the system will be as follows:
$W=\dfrac{R\times 30}{\dfrac{5}{2}-1}$
$\Rightarrow W=\dfrac{60R}{3}$
$\Rightarrow W=20R$
Hence, work done for the given system is $20R$. Thus, option (C) is the correct answer.

Note:
It is important to note that the value of work done for the given system can alternatively be calculated using the direct formula of work done i.e., $dW=-pdV$ and integrating it in terms of temperature to get the final answer. Also, remember that the Poisson’s equation is only applicable for reversible adiabatic processes not for irreversible processes.