Question

In ${{H}_{2}}$ gas process $\text{P}{{\text{V}}^{\text{2}}}\text{= Constant}$.Then ratio of work done by gas to change in its internal energy is
A. $\dfrac{2}{3}$
B. $0.4$
C. $-0.4$
D. $-\dfrac{2}{3}$

Answer 1

Since $\text{P}{{\text{V}}^{\text{2}}}\text{= Constant}$,it is polytropic process. Hence we can use the formula for work done in a polytropic process. Also have equation for internal energy of a diatomic molecule. Using both the formulas, the ratio of work done by hydrogen gas to change in its internal energy can be calculated.
Formula used:
$\text{Work done by the gas, W =}\int\limits_{{{\text{V}}_{\text{1}}}}^{{{\text{V}}_{\text{2}}}}{\text{Pdv}}$
$PV=nRT$
$\text{Internal energy, U=}\dfrac{\text{5}}{\text{2}}\text{nRT}$

Complete answer:
Given,
$P{{V}^{2}}=K$ -------- 1
Where,
$\text{P = Pressure of the system}$
$\text{V = Volume of the gas}$
$\text{K = constant}$
Hence it is a polytropic process.
Then,
$P=\dfrac{K}{{{V}^{2}}}$ ------- 2
For a polytropic process,
$\text{Work done by the gas, W =}\int\limits_{{{\text{V}}_{\text{1}}}}^{{{\text{V}}_{\text{2}}}}{\text{Pdv}}$ --------- 3
Substitute equation 2 in 3. Then,
$W=\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{\dfrac{K}{{{V}^{2}}}dv}=\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{K\dfrac{dv}{{{V}^{2}}}}$
$W=K\left[ \dfrac{-1}{V} \right]_{{{V}_{1}}}^{{{V}_{2}}}=-K\left[ \dfrac{1}{{{V}_{2}}}-\dfrac{1}{{{V}_{1}}} \right]=-\dfrac{K}{{{V}_{2}}}+\dfrac{K}{{{V}_{1}}}$------- 4
Substitute equation 1 in 4. We get,
$W=\dfrac{-{{P}_{2}}V_{2}^{2}}{{{V}_{2}}}+\dfrac{{{P}_{1}}V_{1}^{2}}{{{V}_{1}}}=-{{P}_{2}}{{V}_{2}}+{{P}_{1}}{{V}_{1}}$--------- 5
We have,
$PV=nRT$ ------ 6
Where,
$\text{n = No}\text{. of moles}$
$\text{T = Temperature}$
$\text{R = Ideal gas constant = 8}\text{.314 J/mol}$
Substituting 6 in equation 5, we get,
$W=-nR{{T}_{2}}+nR{{T}_{1}}=-\left( nR{{T}_{2}}-nR{{T}_{1}} \right)$
$W=-nR\Delta T$
For a diatomic molecule,
$\text{Internal energy, U=}\dfrac{\text{5}}{\text{2}}\text{nRT}$
Therefore,
$\text{Change in internal energy, }\\!\\!\Delta\\!\\!\text{ U=}\dfrac{\text{5}}{\text{2}}\text{nR }\\!\\!\Delta\\!\\!\text{ T}$
Then,
$\dfrac{W}{\Delta U}=\dfrac{-nR\Delta T}{\dfrac{5}{2}nR\Delta T}$
$\dfrac{W}{\Delta U}=-\dfrac{2}{5}=-0.4$

So, the correct answer is “Option C”.

Additional Information:
A polytropic process is any thermodynamic process which can be expressed as,
$\text{P}{{\text{V}}^{\text{n}}}\text{ = Constant}$. The polytropic process can be described as the expansion and compression of a gas which includes heat transfer. Here, the exponent $n$ is known as the polytropic index, and it can take up any value from$\text{0 to }\infty$, depending on the process.

Note:
Here in the expression of a polytropic process, n is known as polytropic index. Depending on the process, its values may vary from $\text{0 to }\infty$. As the value of $n$ approaches $\infty$, pressure has less influence; when $\text{n = }\infty$, pressure has no influence. Hence, we have an isobaric process.
For an irreversible process, we cannot use the same formula for calculating the work done, which we use in a reversible process.