A horizontal cylinder with insulating walls is divided into two parts by a frictionless, insulating…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A horizontal cylinder with insulating walls is divided into two parts by a frictionless, insulating piston. At first each part holds $10.0\ \text{litres}$ of a monatomic ideal gas at $1.0 \times 10^{5}\ \text{Pa}$ and $300\ \text{K}$ . A small electric coil in the left part now heats the gas there slowly, as shown in the figure, until the gas in the right part has been squeezed to half its volume.

A horizontal cylinder with insulating walls, divided by a free, insulating piston into a left part and a right part of 10.0 litres each. A small electric coil is in the left part. Both parts hold a monatomic gas at 1.0 times 10 to the 5 Pa and 300 K at the start.

What is the final pressure?
What are the final temperatures in the two parts?
How much work is done on the gas in the right part?
How much heat does the coil supply?

The worked solution

Have you tried it?

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. The right gas is squeezed adiabatically. The left gas has the same pressure and the rest of the volume. Then use the first law for each side.

Step 1. Pressure (a). For the right part, $PV^{5/3}$ stays fixed:

\displaystyle P = P_0\left(\frac{V_0}{V_0/2}\right)^{5/3} = 2^{5/3}P_0 = 3.175 \times 1.0 \times 10^{5}\ \text{Pa} = 3.175 \times 10^{5}\ \text{Pa}

Step 2. Temperatures (b). Right: $T = T_0\,\dfrac{PV}{P_0V_0} = 300 \times 3.175 \times 0.5\ \text{K} = 476.2\ \text{K}$ (that is, $2^{2/3}T_0$ ). Left: the volume is now $15.0\ \text{litres}$ , so $T = 300 \times 3.175 \times 1.5\ \text{K} = 1429\ \text{K}$ .

Step 3. Work on the right gas (c). No heat enters it, so the work done on it is its gain in internal energy:

\displaystyle W = \tfrac32\left(PV_R - P_0V_0\right) = \tfrac32(1587.4 - 1000)\ \text{J} = 881\ \text{J}

Step 4. Heat from the coil (d). The left gas does the same work, $881\ \text{J}$ , on the piston, and its internal energy rises by $\tfrac32(PV_L - P_0V_0) = \tfrac32(4762.2 - 1000)\ \text{J} = 5643\ \text{J}$ . So

\displaystyle Q = 5643 + 881 = 6524\ \text{J}

Answer: (a) $3.17 \times 10^{5}\ \text{Pa}$ . (b) $476.2\ \text{K}$ on the right, $1429\ \text{K}$ on the left. (c) $881\ \text{J}$ . (d) $6524\ \text{J}$ .

Check. The whole cylinder does no work on the outside, so all the coil's heat must appear as internal energy of the two gases: $\tfrac32(4762.2 + 1587.4 - 2 \times 1000)\ \text{J} = 6524\ \text{J}$ . It does.

Watch out. The left gas does not follow an adiabat: it is being heated. Only the right gas is adiabatic. The left gas's state comes from the shared pressure and the leftover volume.

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