A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig. 2.1.

Question 10

(a) (i) State the basic assumption of the kinetic theory of gases that leads to the conclusion that the potential energy between the atoms of an ideal gas is zero. [1]

 

(ii) State what is meant by the internal energy of a substance. [2]

 

(iii) Explain why an increase in internal energy of an ideal gas is directly related to a

rise in temperature of the gas. [2]

 

 

(b) A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig. 2.1.

 

Fig. 2.1

(i) State the change in internal energy of the gas during one complete cycle PQRP. [1]

 

(ii) Calculate the work done on the gas during the change from P to Q. [2]

 

(iii) Some energy changes during the cycle PQRP are shown in Fig. 2.2.

 

Fig. 2.2

Complete Fig. 2.2 to show all of the energy changes. [3]

 

 

Reference: Past Exam Paper – November 2010 Paper 41 & 42 Q2

 

Solution:

(a) (i) There are no forces (of attraction or repulsion) between the atoms / molecules / particles.

 

(ii) The internal energy of a substance is the sum of kinetic and potential energy of the atoms / molecules due to their random motion.

 

(iii) The (random) kinetic energy increases with temperature. There is no potential energy. (since the gas is ideal)

So, an increase in temperature increases the internal energy.

 

 

(b)

(i) Change in internal energy = Zero

{The initial and final states (values of p and V) are identical in a complete cycle, so the change in internal energy is zero.}

 

(ii) Work done = pΔV = 4.0×10⁵ × 6×10^-4 = 240 J

 

(iii)

(values for the change in internal energy should add up to zero.)

{ΔU = ΔQ + ΔW

+ΔQ is the amount of heat/energy supplied to the gas.

+ΔU is the increase in internal energy.

+ ΔW is the work done ON the gas.   ΔW = p ΔV

For P to Q: ΔQ = - 600 J. As calculate above, ΔW = + 240 J. ΔU = - 600 + 240 = - 360 J

For Q to R: ΔQ = +720 J. ΔV = 0, so ΔW = 0. ΔQ = ΔU – 0 = +720 J

 

For R to P: ΔQ = +480 J.

As answered in (b)(i), the change in internal energy during one complete cycle is zero.

Let the change in internal energy for this change be ΔU. Consider the changes in internal energy for P to Q and Q to R.

-360 + 720 + ΔU = 0.

So, ΔU = -360 J.

ΔW = ΔU – ΔQ = -360 – 480 = -840 J

Note that from R to P, both the pressure and the volume are changing. So, we cannot use the simple equation of W = pΔV, which assumes a constant pressure.}

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}