Two rigid, insulated vessels are joined by a thin tube with a closed valve, as shown in the figure…

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Two rigid, insulated vessels are joined by a thin tube with a closed valve, as shown in the figure. Vessel A, of volume $10.0\ \text{litres}$ , holds $2.0\ \text{mol}$ of helium (a monatomic gas) at $300\ \text{K}$ . Vessel B, of volume $20.0\ \text{litres}$ , holds $1.0\ \text{mol}$ of nitrogen (a diatomic gas with rigid molecules) at $600\ \text{K}$ . The valve is opened and the gases mix. Treat both as ideal gases, ignore the volume of the tube and the heat capacity of the vessels, and take $R = 8.314\ \text{J/(mol K)}$ .

Two closed vessels joined by a thin tube with a closed valve in the middle. Vessel A, on the left, is 10.0 litres and holds 2.0 mol of helium at 300 K. Vessel B, on the right, is 20.0 litres and holds 1.0 mol of nitrogen at 600 K. Both vessels are insulated.

Find the pressure in each vessel before the valve is opened.
What is the final temperature of the mixture?
What is the final pressure?
How much heat must then be taken out of the mixture, at constant volume, to bring it back to $300\ \text{K}$ ?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. No work and no heat, so the internal energy stays the same. Find the temperature from that, then the pressure from the gas law.

Step 1. Before (a). $P_A = \dfrac{n_ART_A}{V_A} = \dfrac{2.0 \times 8.314 \times 300}{1.0 \times 10^{-2}}\ \text{Pa} = 4.988 \times 10^{5}\ \text{Pa}$ , and likewise $P_B = 2.494 \times 10^{5}\ \text{Pa}$ .

Step 2. Energy (b). In units of $R$ : helium has $\tfrac32 \times 2.0 \times 300 = 900R$ and nitrogen $\tfrac52 \times 1.0 \times 600 = 1500R$ , so $2400R$ in all. The total heat capacity is $(\tfrac32 \times 2.0 + \tfrac52 \times 1.0)R = 5.5R$ . With $U$ unchanged,

\displaystyle T = \frac{2400}{5.5}\ \text{K} = 436.4\ \text{K}

Step 3. Pressure (c). $3.0\ \text{mol}$ now fill $30.0\ \text{litres}$ :

\displaystyle P = \frac{3.0 \times 8.314 \times 436.4}{3.0 \times 10^{-2}}\ \text{Pa} = 3.628 \times 10^{5}\ \text{Pa}

(helium $2.419 \times 10^{5}\ \text{Pa}$ plus nitrogen $1.209 \times 10^{5}\ \text{Pa}$ ).

Step 4. Cooling back (d). At constant volume, $Q = C_V\,\Delta T = 5.5 \times 8.314 \times 136.4\ \text{J} = 6236\ \text{J}$ must be taken out.

Answer: (a) $4.99 \times 10^{5}\ \text{Pa}$ in A, $2.49 \times 10^{5}\ \text{Pa}$ in B. (b) $436.4\ \text{K}$ . (c) $3.63 \times 10^{5}\ \text{Pa}$ . (d) $6236\ \text{J}$ .

Check. Think of it as an exchange of heat: the nitrogen cools by $163.6\ \text{K}$ and gives out $\tfrac52 R \times 163.6 = 3401\ \text{J}$ ; the helium warms by $136.4\ \text{K}$ and takes in $3R \times 136.4 = 3401\ \text{J}$ . They agree.

Watch out. The final temperature is not the average of the two weighted by moles alone. Weight by heat capacities: a mole of nitrogen holds more energy per kelvin than a mole of helium.

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The worked solution

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