Answer
See the detailed answer below.
Work Step by Step
a)
We are given, for both gases, that:
- $n=0.10$ mol.
- $T_1=300$ K.
- $P_1=3$ atm.
- $P_2=1$ atm.
- $\gamma=1.4$ since it is a diatomic gas.
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$\rightarrow$ Gas A undergoes an isothermal process, so $T_1=T_{2A}$,
Therefore,
$$T_{2A}=\color{red}{\bf 300}\;\rm K$$
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$\rightarrow$ Gas B undergoes an adiabtic process, so
$$P_1V_1^\gamma=P_{2}V^\gamma_{2B} $$
Hence,
$$\dfrac{P_1}{P_2}=\dfrac{V^\gamma_{2B} }{V_1^\gamma}=\left[\dfrac{V_{2B}}{V_1}\right]^\gamma$$
$$\left[\dfrac{P_1}{P_2}\right]^\frac{1}{\gamma}=\dfrac{V_{2B}}{V_1} $$
Therefore,
$$V_{2B}=V_1\left[\dfrac{P_1}{P_2}\right]^\frac{1}{\gamma} \tag 1$$
Now we need to find the initial volume of the two gases;
$$P_1V_1=nRT_1$$
$$V_1=\dfrac{nRT_1}{P_1}$$
Plugging the known;
$$V_1=\dfrac{(0.1)(8.31)(300)}{(3\times 1.013\times 10^5)}$$
$$V_1=\bf8.20\times 10^{-4} \;\rm m^3=\bf 820\;\rm cm^3$$
Plugging the known into (1);
$$V_{2B}=820\left[\dfrac{3}{1}\right]^\frac{1}{1.4} $$
$$V_{2B}= \bf 1797\;\rm cm^3$$
Now the final temperature of this process is given by
$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_{2B}}{T_{2B}}$$
Hence,
$$T_{2B}=\dfrac{ P_2V_{2B}T_1}{ P_1V_1} $$
Plugging the known;
$$T_{2B}=\dfrac{ (1)(1797)(300)}{ (3)(820)} $$
$$T_{2B}=\color{red}{\bf 219}\;\rm K$$
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b) First, we need to find $V_{2A}$ which is given by
$$V_{2A}=\dfrac{P_1V_1}{P_2}=\dfrac{(3)(820)}{(1)}=\bf 2460\;\rm cm^3$$
$\bullet$ Both gases start from the same point:
$(V_1,P_1)=\rm (820\;cm^3,\;3\;atm)$
$\Rightarrow$ Gas A ends at the point of $(V_{2A},P_{2A})=\rm (2460\;cm^3,\;1\;atm)$
$\Rightarrow$ Gas B ends at the point of $(V_{2B},P_{2B})=\rm (1797\;cm^3,\;1\;atm)$
See the graph below.
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