Question

An ideal diatomic gas (\[\gamma = \dfrac{5}{7}\]​) undergoes a process in which its internal energy relates to the volume as \[U = \alpha V\]​, where α is a constant. If the work (in J) performed by the gas to increase its internal energy by 100J is 16k.Then what is the value of k?

Answer 1

Hint: \[\gamma = \dfrac{{{C_p}}}{{{C_v}}}\]is the formula required to find the given answer. Place the values correctly from the question in the formula to find the answer.
Diatomic molecules are molecules composed of only two atoms, of the same or different chemical elements. If a diatomic molecule consists of two atoms of the same element, such as hydrogen or oxygen, then it is said to be homonuclear.

Complete step by step solution:
First write down all the values which are provided,
\[\gamma = \dfrac{7}{5}\]
\[U = \alpha \sqrt V \]
\[{U_2} - {U_1} = 100\] and
W= 16k

Now from the formula we know that,
\[\gamma = \dfrac{{{C_p}}}{{{C_v}}}\]
So we know the value of \[\gamma \] then we can put the value of \[\gamma \]in \[\gamma = \dfrac{{{C_p}}}{{{C_v}}}\]and can find the expression for \[{C_p}\],
\[\dfrac{7}{5} = \dfrac{{{C_p}}}{{{C_v}}}\]
Now,
\[{C_p} = \dfrac{7}{5}{C_v}\] after simplification
We know that the formula for R is in terms of \[{C_p}\] and \[{C_v}\]is ,
\[{C_p} - {C_v} = R\]

Now putting the value of \[{C_p}\] in the following equation then we get,
\[\dfrac{7}{5}{C_v} - {C_v} = R\]
Solving it we get,
\[\dfrac{{7{C_v} - 5{C_v}}}{5} = R\]
Further simplifying it we get,
\[\dfrac{2}{5}{C_v} = R\]
Now arranging it properly we get,
\[{C_v} = \dfrac{5}{2}R\]
We know that,
 \[U = n{C_v}T\] and in the question value of U is also given as \[U = \alpha \sqrt V \]

Now equating we get,
\[n{C_v}T = \alpha \sqrt V \]

Now putting the value of \[{C_v}\] in the above equation we get,
\[n\dfrac{5}{2}RT = \alpha \sqrt V \]
Further simplifying it we get,
\[nRT = \dfrac{2}{5}\alpha \sqrt V \]
We know that
\[{U_2} - {U_1} = 100\]
So, we can also write,
\[\alpha (\sqrt {{V_2}} - \sqrt {{V_1}} ) = 100\]since \[U = \alpha \sqrt V \]

Now integrating we get,
\[W = \int\limits_{{V_1}}^{{V_2}} {Pdv} \]
We know that \[PV = nRT\] and \[P = \dfrac{{nRT}}{V}\] so putting the value of P in the above equation we get,
\[W = \int\limits_{{V_1}}^{{V_2}} {\dfrac{{nRT}}{V}dv} \]
Now putting the value of nRT which we got is,
\[W = \int\limits_{{V_1}}^{{V_2}} {\dfrac{{\dfrac{2}{5}\alpha \sqrt V }}{V}dv} \]
Further integrating it we get,
\[W = \dfrac{4}{5}\alpha (\sqrt {{V_2}} - \sqrt {{V_1}} )\]
Now putting the value of \[\alpha (\sqrt {{V_2}} - \sqrt {{V_1}} ) = 100\]we get,
\[W = \dfrac{4}{5} \times 100\]
Which is
W= 80J
From the given question we know that W=16k so putting the value of W we can get the value of k,
80=16k
k=5

Therefore, the value of k=5

Note: Students need to be able to correctly equate the formulas for the gases. They should also be able to intuitively simplify the equations whenever required. Students often go wrong in a few of the first steps. They put the wrong values in the equation so they fail to get the right answer.