HardPhysicsMiddle Stage (6–8)Heat & Thermodynamics › Thermodynamic Processes and Specific Heats of GasesNumericalJEE Advanced JEE Main

A vertical cylinder, closed at the bottom, holds a monatomic ideal gas under a light, frictionless…

Contributed by StemPy Lab

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

A vertical cylinder, closed at the bottom, holds a monatomic ideal gas under a light, frictionless piston of area $0.010\ \text{m}^2$ . Above the piston is the open air, at $1.0 \times 10^{5}\ \text{Pa}$ . A spring of force constant $1.0 \times 10^{4}\ \text{N/m}$ joins the top of the piston to a fixed support, as shown in the figure. At first the spring has its natural length, and the gas has a volume of $1.0\ \text{litre}$ at $300\ \text{K}$ . The gas is now heated slowly until the piston has risen by $10\ \text{cm}$ . Take $R = 8.314\ \text{J/(mol K)}$ .

A vertical cylinder, closed at the bottom, holds gas under a light piston of area 0.010 square m. The space above the piston is open to the air at 1.0 times 10 to the 5 Pa. A spring of force constant 1.0 times 10 to the 4 N/m joins the top of the piston to a fixed support above the cylinder.

What are the final pressure and the final temperature of the gas?
How much work does the gas do? How much of it goes into the spring?
How much heat is given to the gas?
What is the molar heat capacity of the gas in this process, in $\text{J/(mol K)}$ ? Write it as a multiple of $R$ .

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. Find the pressure from the forces on the piston, the work from the area under the path, and the heat from the first law.

Step 1. Final state (a). The spring is squeezed by $10\ \text{cm}$ and pushes down with $kx$ , so

\displaystyle P_2 = P_0 + \frac{kx}{A} = 1.0 \times 10^{5}\ \text{Pa} + 1.0 \times 10^{5}\ \text{Pa} = 2.0 \times 10^{5}\ \text{Pa}

The volume grows by $Ax$ to $2.0\ \text{litres}$ . From $PV/T$ fixed: $T_2 = 300\ \text{K} \times \dfrac{2.0 \times 10^{5} \times 2.0}{1.0 \times 10^{5} \times 1.0} = 1200\ \text{K}$ .

Step 2. Work (b). The pressure grows in a straight line from $1.0 \times 10^{5}$ to $2.0 \times 10^{5}\ \text{Pa}$ , so the area under the path is a trapezium:

\displaystyle W = \frac{P_0 + P_2}{2}\,\Delta V = 150\ \text{J}

Of this, $P_0 A x = 100\ \text{J}$ pushes back the air and $\tfrac12 kx^2 = 50\ \text{J}$ is stored in the spring.

Step 3. Heat (c). $\Delta U = \tfrac32(P_2V_2 - P_1V_1) = \tfrac32(400 - 100)\ \text{J} = 450\ \text{J}$ , so $Q = \Delta U + W = 450 + 150 = 600\ \text{J}$ .

Step 4. Heat capacity (d). The amount of gas is $n = \dfrac{P_1V_1}{RT_1} = 0.0401\ \text{mol}$ , and $n\Delta T = 36.08\ \text{mol K}$ . So

\displaystyle C = \frac{Q}{n\Delta T} = \frac{600}{36.08}\ \text{J/(mol K)} = 16.6\ \text{J/(mol K)} = 2.0R

Answer: (a) $2.0 \times 10^{5}\ \text{Pa}$ , $1200\ \text{K}$ . (b) $150\ \text{J}$ , of which $50\ \text{J}$ goes into the spring. (c) $600\ \text{J}$ . (d) $16.6\ \text{J/(mol K)}$ , that is $2R$ .

Check. The heat capacity lies between $C_V = 1.5R$ (no work) and $C_P = 2.5R$ (pressure fixed), as it should for a path on which the pressure rises a little as the gas expands. A quick way to the same value: $nR\Delta T = P_2V_2 - P_1V_1 = 300\ \text{J}$ , so $\dfrac{C}{R} = \dfrac{Q}{nR\Delta T} = \dfrac{600}{300} = 2$ .

Watch out. Do not use $W = P\Delta V$ with one pressure: the pressure changes along the way. And remember the work against the air: it is the larger part here.

← All Problems

← At very low temperatures, the heat capacity of a crystal grows as the cube of its temperature (in…Two rigid, insulated vessels are joined by a thin tube with a closed valve, as shown in the figure. →

More onHeat & Thermodynamics › Thermodynamic Processes and Specific Heats of Gases

Discussion (0)

Loading…