
Ideal monatomic gas is taken through a process $dQ=2dU$. What will be the molar heat capacity for the process? (where $dQ$ is heat supplied and $dU$ is change in internal energy)
$\begin{align}
& A.2R \\
& B.3R \\
& C.4R \\
& D.5R \\
\end{align}$
Answer
566.1k+ views
Hint: The change in heat energy at the constant pressure is given as the product of number of moles, specific heat at constant volume and the change in temperature. The change in internal energy is given as the product of the number of moles, specific heat at constant pressure and the change in temperature. Equate both the equations according to the relation and find the answer.
Complete step by step answer:
the change in heat energy at the constant pressure is given as
$dQ=n{{C}_{p}}dT$
Where \[{{C}_{p}}\] be the specific heat at constant pressure, \[n\] be the number of moles and \[dT\] be the change in temperature.
The change in internal energy at constant volume is given as,
\[dU=n{{C}_{v}}dT\]
Where \[{{C}_{v}}\] be the specific heat at constant volume.
As per the question, the change in heat energy is equal to the twice of the change in internal energy.
That is,
\[dQ=2dU\]
Substituting the equations in it will give,
\[n{{C}_{p}}dT=2n{{C}_{v}}dT\]
We can cancel the common terms on both the sides. Therefore we can write that,
\[{{C}_{p}}=2{{C}_{v}}\]
The specific heat at constant volume is given by the formula,
\[{{C}_{v}}=\dfrac{R}{\gamma -1}\]
Where \[\gamma \]be the Poisson’s ratio.
Substituting the value in the equation will give,
\[{{C}_{p}}=2\dfrac{R}{\gamma -1}\]
For monoatomic gases, the Poisson’s ratio will be,
\[\gamma =\dfrac{5}{3}\]
Substituting the value in the equation will give,
\[\begin{align}
& {{C}_{P}}=\dfrac{2R}{\dfrac{5}{3}-1}=\dfrac{2R}{\dfrac{2}{3}}=3R \\
& {{C}_{p}}=3R \\
\end{align}\]
So, the correct answer is “Option B”.
Note: Poisson’s ratio is the measure of the Poisson’s phenomenon. It explains the contraction and elongation of a body. It can be expressed as the ratio of the specific heat capacity at constant pressure to the specific heat capacity at constant volume.
Complete step by step answer:
the change in heat energy at the constant pressure is given as
$dQ=n{{C}_{p}}dT$
Where \[{{C}_{p}}\] be the specific heat at constant pressure, \[n\] be the number of moles and \[dT\] be the change in temperature.
The change in internal energy at constant volume is given as,
\[dU=n{{C}_{v}}dT\]
Where \[{{C}_{v}}\] be the specific heat at constant volume.
As per the question, the change in heat energy is equal to the twice of the change in internal energy.
That is,
\[dQ=2dU\]
Substituting the equations in it will give,
\[n{{C}_{p}}dT=2n{{C}_{v}}dT\]
We can cancel the common terms on both the sides. Therefore we can write that,
\[{{C}_{p}}=2{{C}_{v}}\]
The specific heat at constant volume is given by the formula,
\[{{C}_{v}}=\dfrac{R}{\gamma -1}\]
Where \[\gamma \]be the Poisson’s ratio.
Substituting the value in the equation will give,
\[{{C}_{p}}=2\dfrac{R}{\gamma -1}\]
For monoatomic gases, the Poisson’s ratio will be,
\[\gamma =\dfrac{5}{3}\]
Substituting the value in the equation will give,
\[\begin{align}
& {{C}_{P}}=\dfrac{2R}{\dfrac{5}{3}-1}=\dfrac{2R}{\dfrac{2}{3}}=3R \\
& {{C}_{p}}=3R \\
\end{align}\]
So, the correct answer is “Option B”.
Note: Poisson’s ratio is the measure of the Poisson’s phenomenon. It explains the contraction and elongation of a body. It can be expressed as the ratio of the specific heat capacity at constant pressure to the specific heat capacity at constant volume.
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