Question 101: Thermodynamics & KTG

Question

One gas of $n_1$ mole of molecules at temperature $T_1$, volume $V_1$, and pressure $P_1$, and another gas of $n_2$ mole of molecules at temperature $T_2$, volume $V_2$, and pressure $P_2$, are mixed resulting in pressure $P$ and volume $V$ of the mixture. The temperature of the mixture is $\_\_\_\_$ .

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

$$ \left(\mathrm{T}_1+\mathrm{T}_2\right) / 2 $$

Answer

$$ \mathrm{T}_1 \mathrm{~T}_2 \mathrm{PV} /\left(\mathrm{T}_2 \mathrm{P}_1 \mathrm{~V}_1+\mathrm{T}_1 \mathrm{P}_2 \mathrm{~V}_2\right) $$

Answer

$$ \left(\mathrm{T}_2 \mathrm{P}_1 \mathrm{~V}_1+\mathrm{T}_1 \mathrm{P}_2 \mathrm{~V}_2\right) /\left(\mathrm{T}_1 \mathrm{~T}_2 \mathrm{PV}\right) $$

Answer

$$ \left|\mathrm{T}_1-\mathrm{T}_2\right| / 2 $$

Thermodynamics & KTG

One gas of n_1 mole of molecules at temperature T_1, volume V_1, and pressure P_1, and another gas of n_2 mole of molecules at temperature T_2, volume V_2, and pressure P_2, are mixed resulting in pressure P and volume V of the mixture. The temperature of the mixture is \_\_\_\_ .

Choose the correct answer: