Question 102: Thermodynamics & KTG

Question

An ideal gas undergoes a process maintaining relation between pressure $(P)$ and $\operatorname{volume}(V)$ as $P=P_{\mathrm{o}}\left(1+\left(\frac{V_{\mathrm{o}}}{V}\right)^2\right)^{-1}$, where $P_{\mathrm{o}}$ and $V_{\mathrm{o}}$ are constants. If two samples $A$ and $B$ (two moles each) with initial volumes $V_{\mathrm{o}}$ and $3 V_{\mathrm{o}}$ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, $T_B-T_A$ is $\_\_\_\_$ .
( $R=$ gas constant)

Accepted Answer

For the given process,
$$ P = P_0\left(1+\left(\frac{V_0}{V}ight)^2ight)^{-1} = \frac{P_0}{1+\left(\frac{V_0}{V}ight)^2} $$
Both samples contain $2$ moles of ideal gas, so at any stage,
$$ PV = nRT = 2RT $$
Hence,
$$ T = \frac{PV}{2R} $$
We are told that the two samples attain the same pressure during the process.
So first let us find the volume of each sample when the pressure is equal.
Since pressure depends only on $V$ through the given relation,
$$ P = \frac{P_0}{1+\left(\frac{V_0}{V}ight)^2} $$
for a given pressure, the corresponding volume is uniquely fixed. Therefore, if both samples attain the same pressure, their final volumes must be the same.
Let this common final volume be $V$.
Then their temperatures at that pressure will be
$$ T = \frac{PV}{2R} $$
So if both have same $P$ and same $V$, their temperatures would also be same.
But the question is asking for $T_B - T_A$, which indicates we should compare the temperatures of the two samples at their given initial volumes under the same process relation.
So let us calculate temperature for each sample using the given $P$-$V$ relation.
Using
$$ T = \frac{PV}{2R} $$
and
$$ P = \frac{P_0}{1+\left(\frac{V_0}{V}ight)^2} $$
we get
$$ T = \frac{P_0 V}{2R\left(1+\left(\frac{V_0}{V}ight)^2ight)} $$
Now for sample $A$:
Initial volume,
$$ V_A = V_0 $$
So pressure is
$$ P_A = \frac{P_0}{1+\left(\frac{V_0}{V_0}ight)^2} = \frac{P_0}{1+1} = \frac{P_0}{2} $$
Thus,
$$ T_A = \frac{P_A V_A}{2R} = \frac{\left(\frac{P_0}{2}ight)V_0}{2R} = \frac{P_0V_0}{4R} $$
Now for sample $B$:
Initial volume,
$$ V_B = 3V_0 $$
So pressure is
$$ P_B = \frac{P_0}{1+\left(\frac{V_0}{3V_0}ight)^2} = \frac{P_0}{1+\frac{1}{9}} = \frac{P_0}{\frac{10}{9}} = \frac{9P_0}{10} $$
Thus,
$$ T_B = \frac{P_B V_B}{2R} = \frac{\left(\frac{9P_0}{10}ight)(3V_0)}{2R} = \frac{27P_0V_0}{20R} $$
Now,
$$ T_B - T_A = \frac{27P_0V_0}{20R} - \frac{P_0V_0}{4R} $$
Taking LCM:
$$ T_B - T_A = \frac{27P_0V_0 - 5P_0V_0}{20R} = \frac{22P_0V_0}{20R} = \frac{11P_0V_0}{10R} $$
Therefore,
$$ \boxed{T_B - T_A = \frac{11P_0V_0}{10R}} $$
So the correct option is:
$$ \boxed{	ext{Option B}} $$

Answer

$\frac{9 P_{\mathrm{o}} V_{\mathrm{o}}}{8 R}$

Answer

$\frac{11 P_{\mathrm{o}} V_{\mathrm{o}}}{10 R}$

Answer

$$ \frac{7 P_{\mathrm{o}} V_{\mathrm{o}}}{6 R} $$

Answer

$$ \frac{13 P_{\mathrm{o}} V_{\mathrm{o}}}{11 R} $$

Thermodynamics & KTG

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