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.}