CIE AS & A Level Physics : 15.1 The mole – Exam style question – Paper 4

Question

A fixed mass of an ideal gas has a volume V and a pressure p. The gas undergoes a cycle of changes, X to Y to Z to X, as shown in Fig. 2.1.

Table 2.1 shows data for p, V and temperature T for the gas at points X, Y and Z.

Table 2.1

(a)     State the change in internal energy ΔU for one complete cycle, XYZX.

(b)     Calculate the amount n of gas.

(c)     Complete Table 2.1.

Use the space below for any working.

(d)     (i)     The first law of thermodynamics for a system may be represented by the equation

ΔU = q + W.

State, with reference to the system, what is meant by:

(ii)    Explain how the first law of thermodynamics applies to the change Z to X.

Ans:

(a)      $$0$$

(b)     $$pV$$ = $$nRT$$

$$(n =) 1.5 \times 10^{5} \times 4.2 \times 10^{–3} / 8.31 \times 540$$

= 0.14 mol

(c)     missing pressure $$1.5 ( \times 10^{5})$$

both missing volumes  $$1.8 ( \times 10^{–3})$$

(d)     (i)      $$(ΔU:)$$ increase in internal energy (of the system)

$$(q:)$$ thermal energy supplied to the system B1

$$(W:)$$ work done on system

(ii)      volume increases and work is done by the gas

Question

A fixed mass of an ideal gas is at a temperature of 21°C. The pressure of the gas is 2.3 × 105 Pa and its volume is 3.5 × 10–3m3.

(a) (i) Calculate the number N of molecules in the gas.
N = ………………………………………………… [2]

(ii) The mass of one molecule of the gas is 40u.
Determine the root-mean-square (r.m.s.) speed of the gas molecules.
r.m.s. speed = ………………………………………… ms–1 [2]

(b) The temperature of the gas is increased by 84°C.
Calculate the value of the ratio

ratio = ………………………………………………… [2]
[Total: 6]

Ans

(a) (i) pV =  NkT = or  pV = nRT = and N = nNA

$$N=\frac{2.3\times 10^{5}\times 3.5\times 10^{-3}}{1.38\times 10^{-23}\times 294}$$

= 2.0 × 1023

(a) (ii) $$pV=\frac{1}{3}Nmc^{2}$$

$$c^{2}=\frac{3\times 2.3\times 10^{5}\times 3.5\times 10^{-3}}{2.0\times 10^{23}\times 40\times 1.66^{-27}}$$

= 182 000
r.m.s. speed = 430 ms–1

or

$$\frac{1}{2}mc^{2}=\frac{3}{2}kT$$

$$c^{2}=\frac{3\times 1.38\times 10^{-23}\times 294}{40\times 1.66\times 10^{-27}}$$

= 183 000
r.m.s.speed = 430 ms–1

(b)  $$c^{2}=\frac{3\times2.0\times10^{23}\times1.38\times 10^{-23}\times(294+84)}{2.0\times 10^{23}\times 40\times 1.66\times 10^{-27}}$$

c = 236000

c= 485

$$ratio=\left ( =\frac{485}{430} \right )=1.1$$

OR

$$v\infty \sqrt{T} \ or\ v^{2}\infty T$$

$$ratio=\sqrt{\frac{273+21+84}{273+21}} \ or\ \sqrt{\frac{378}{294}}$$

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