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

Hint: 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.