At point A, the gas has volume 2.4 × 10-3 m3, pressure 1.6 × 105 Pa and temperature 300 K.

Question 9

(a) The first law of thermodynamics may be expressed in the form

ΔU = q + w.

(i) State, for a system, what is meant by:

1. +q

2. +w.

[2]

(ii) State what is represented by a negative value of ΔU. [1]

(b) An ideal gas, sealed in a container, undergoes the cycle of changes shown in Fig. 2.1.

Fig. 2.1

At point A, the gas has volume 2.4 × 10^-3 m³, pressure 1.6 × 10⁵ Pa and temperature 300 K.

The gas is compressed suddenly so that no thermal energy enters or leaves the gas during the compression. The amount of work done is 480 J so that, at point B, the gas has volume 8.7 × 10^-4 m³, pressure 6.6 × 10⁵ Pa and temperature 450 K.

The gas is now cooled at constant volume so that, between points B and C, 1100 J of thermal energy is transferred. At point C, the gas has pressure 1.6 × 10⁵ Pa and temperature 110 K.

Finally, the gas is returned to point A.

(i) State and explain the total change in internal energy of the gas for one complete

cycle ABCA. [2]

(ii) Calculate the external work done on the gas during the expansion from point C to

point A. [2]

(iii) Complete Fig. 2.2 for the changes from:

1. point A to point B

2. point B to point C

3. point C to point A.

Fig. 2.2

[4]

[Total: 11]

Reference: Past Exam Paper – June 2019 Paper 42 Q2

Solution:

(a)

(i)

1. energy transfer to the system by heating

2. (external) work done on the system

(ii) decrease in internal energy

(b)

(i) The initial and final temperatures are the same, hence the net change in internal energy around a full cycle is zero.

(ii)

work done = pΔV

work done = (–)1.6 × 10⁵ × (2.4 – 0.87)×10^-3

work done = (–)240 J

(iii)

{As given in the question,

From A to B,

the gas is compressed suddenly so that no thermal energy enters or leaves. (i.e. +q = 0)

the gas is compressed, so work is done ON the gas. +w = 480 J

ΔU = q + w = 0 + 480 = 480 J

From B to C,

the volume of the gas does not change (as seen in the graph), so ΔV = 0 and w = p ΔV = 0

the gas is cooled, so it loses thermal energy. q = - 1100 J

ΔU = q + w = -1100 + 0 = -1100 J

From C to A,

the gas expands, so it does work. w is negative. As calculated previously, w = - 240 J

in (b)(i), we have stated that the total change in internal energy in a closed cycle is zero.

let the change in internal energy from C to A be U.

(adding values in the last column should equal to 0)

U + 480 – 1100 = 0

U = 620 J

ΔU = q + w

620 = q – 240

q = 620 + 240 = 860 J}