AP Physics 2- 9.3 Thermal Energy Transfer and Equilibrium- Exam Style questions - FRQs- New Syllabus

(a)(i)
Place the cylinder in a large water bath so that the part of the container below the piston is submerged. The water bath helps keep the gas at a constant temperature because thermal energy can be transferred between the gas and the water bath until thermal equilibrium is maintained.

Use the meterstick to measure the radius \(r\) of the cylindrical piston. The students could also measure the diameter and use \(r=\dfrac{D}{2}\). Measure the height \(h\) of the gas column from the bottom of the container to the piston.

First, record the height \(h\) with only the piston on the gas. Then place \(1\) object of known mass \(m_o\) on top of the piston and wait for the piston to stop moving. Record the new height \(h\). Repeat the process using \(2\), \(3\), \(4\), and more objects so that more than two data points are collected.

For each trial, the added objects increase the downward force on the piston, which increases the pressure of the gas. The volume of the gas is found from the cylinder geometry using \(V=\pi r^2 h\). The density can then be calculated using \(\rho=\dfrac{M_g}{V}\).

To improve consistency, the students should wait after each added object until the piston stops moving and the gas returns to the temperature of the water bath before recording the height.

(a)(ii)
The absolute pressure of the gas must balance the atmospheric pressure plus the pressure caused by the weight of the piston and the added objects.

The area of the piston is \(A=\pi r^2\).

If \(N\) objects are placed on the piston, the total downward weight on the gas due to the piston and objects is \((m_p+Nm_o)g\).

Therefore, the additional pressure due to the piston and objects is \(\dfrac{(m_p+Nm_o)g}{\pi r^2}\).

The absolute pressure of the gas is

\(\boxed{P_{\text{gas}}=P_{\text{atm}}+\dfrac{(m_p+Nm_o)g}{\pi r^2}}\)

where \(P_{\text{atm}}\) is the atmospheric pressure, \(N\) is the number of objects placed on the piston, \(m_p\) is the mass of the piston, \(m_o\) is the mass of each added object, \(g\) is the gravitational field strength, and \(r\) is the radius of the piston.

(a)(iii)
Density is mass divided by volume.

The mass of the gas is \(M_g\), and the volume of the gas in the cylindrical container is

\(V=\pi r^2h\)

Therefore, the density of the gas is

\(\boxed{\rho=\dfrac{M_g}{\pi r^2 h}}\)

where \(h\) is the measured height of the gas column below the piston.

(a)(iv)
No, the data does not indicate that the gas is ideal.

For an ideal gas, \(PV=nRT\). Since the mass of the gas is constant, the number of moles \(n\) is constant. The experiment is done at constant temperature, so \(T\) is also constant.

This means \(PV\) should be constant.

Since density is \(\rho=\dfrac{M_g}{V}\), the volume is \(V=\dfrac{M_g}{\rho}\).

Substitute this into the ideal gas equation:

\(P\left(\dfrac{M_g}{\rho}\right)=nRT\)

Rearranging gives

\(\rho=\dfrac{M_g}{nRT}P\)

Since \(M_g\), \(n\), \(R\), and \(T\) are constants, the ideal gas model predicts that density should be directly proportional to pressure.

Therefore, a graph of density as a function of pressure should be linear and pass through the origin for an ideal gas at constant temperature.

The given graph is curved, not linear. Since the graph does not show a linear relationship between density and pressure, the data does not support the conclusion that the gas is ideal.

\(\boxed{\text{The gas does not behave ideally because } \rho \text{ is not directly proportional to } P.}\)