AP Physics B – Answer to Free Response Practice Question on Thermodynamics

“Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time”

– Thomas A. Edison

A free response practice question on thermodynamics was posted on 24^th October 2010 for AP Physics B aspirants. As promised, I give below a model answer (along with the question) for your benefit.

The adjoining figure shows a fixed cylindrical vessel of inner diameter 10 cm fitted with a smooth, light piston connected to a light spring of spring constant 4000 Nm^–1. The other end of the spring is attached to an immovable support. The cylinder contains an ideal gas at 27º C. The spring is initially in its released condition and the initial volume of the gas in the cylinder is 0.8×10^–3 m³. When heat is supplied to the gas, it expands and pushes the piston through 10 cm. The atmospheric pressure is 10⁵ pascal. Now answer the following questions:

(a) What are the initial pressure and the initial absolute (Kelvin) temperature of the gas?

(b) Calculate the potential energy acquired by the spring because of the expansion of the gas.

(c) Calculate the final pressure of the gas in the cylinder.

(d) Calculate the final temperature of the gas in the cylinder.

(a) The initial pressure P₁ of the gas is the same as the atmospheric pressure 10⁵ Nm^–1. The initial temperature T₁ of the gas is (27 + 273) K = 300 K.

(b) The potential energy of the spring is U = ½ kx² where k is the spring constant and x is the compression of the spring.

Therefore, U = ½ ×4000×(0.1)² = 20 J.

(c) The final pressure of the gas is the sum of the atmospheric pressure and the pressure due to the elastic force developed in the spring.

Elastic force = kx.

Since this force acts over the area A of the piston, pressure due to the elastic force is kx/A = kx/πR² where R is the radius of the piston.

Therefore, final pressure P₂ of the gas = 10⁵ + (4000×0.1)/(π×0.05²) = 10⁵ + 0.51×10⁵ = 1.51×10⁵ pascal.

(d) The final temperature T₂ of the gas is given by

P₁V₁/T₁ = P₂V₂/T₂

Therefore, T₂ = (P₂V₂ T₁)/ (P₁V₁)

The final volume V₂ of gas = Initial volume + πR²x = 0.8×10^–3 + (π×0.05²×0.1) = 0.8×10^–3 + 0.79×10^–3 = 1.59×10^–3 m³

Therefore, final temperature T₂ = (1.51×10⁵×1.59×10^–3 ×300)/(10⁵×0.8×10^–3) = 900 K.