Question 97: Thermodynamics & KTG

Question

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason $\mathbf{R}$
Statement I: Change in internal energy of a system containing $n$ mole of ideal gas can be written as $\Delta \mathrm{U}=n \mathrm{C}_v\left(T_{\mathrm{f}}-T_i\right)=\frac{n R}{\gamma-1}\left(T_{\mathrm{f}}-T_i\right)$, where $\gamma=\frac{C_p}{C_v}, T_i=$ initial temperature, $T_{\mathrm{f}}=$ final temperature.
Statement II: Relation between degree of freedom $f$ and $\gamma\left(=C_p / C_v\right)$ is $\left(\gamma=1+\frac{2}{f}\right)$
Choose the correct answer from the options given below

Accepted Answer

For an ideal gas, the change in internal energy is
$$ \Delta U = n C_v (T_f - T_i) $$
Now, we use the relation
$$ \gamma=\frac{C_p}{C_v} $$
and also
$$ C_p-C_v=R $$
From these two relations,
$$ \gamma C_v - C_v = R $$
$$ C_v(\gamma-1)=R $$
$$ C_v=\frac{R}{\gamma-1} $$
Substituting in the expression of $\Delta U$,
$$ \Delta U=n\left(\frac{R}{\gamma-1}ight)(T_f-T_i) $$
So Statement I is true.
Now check Statement II.
For an ideal gas with $f$ degrees of freedom,
$$ C_v=\frac{f}{2}R $$
and
$$ C_p=C_v+R=\frac{f}{2}R+R=\left(\frac{f}{2}+1ight)R $$
Hence,
$$ \gamma=\frac{C_p}{C_v} =\frac{\left(\frac{f}{2}+1ight)R}{\frac{f}{2}R} =1+\frac{2}{f} $$
So Statement II is also true.
But Statement II is not the direct explanation of Statement I.
Statement I follows mainly from:
$$ \Delta U=nC_v\Delta T \quad 	ext{and} \quad C_p-C_v=R $$
The relation
$$ \gamma=1+\frac{2}{f} $$
is true, but it is not required to explain Statement I.
Therefore, the correct option is:
$$ \boxed{	ext{Option B}} $$

Answer

Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$

Answer

Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is NOT the correct explanation of $\mathbf{A}$

Answer

A is true but $\mathbf{R}$ is false

Answer

A is false but $\mathbf{R}$ is true

Thermodynamics & KTG

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