An ideal gas initially has pressure 1.0 × 105 Pa, volume 4.0 × 10-4 m3 and temperature 300 K, as illustrated in Fig. 2.1.

Question 5

An ideal gas initially has pressure 1.0 × 10⁵ Pa, volume 4.0 × 10^-4 m³ and temperature 300 K, as illustrated in Fig. 2.1.

Fig. 2.1

A change in energy of the gas of 240 J results in an increase of pressure to a final value of 5.0 × 10⁵ Pa at constant volume.

The thermodynamic temperature becomes T.

(a) Calculate

(i) the temperature T, [2]

(ii) the amount of gas. [2]

(b) The increase in internal energy ΔU of a system may be represented by the expression

ΔU = q + w.

(i) State what is meant by the symbol

1. +q,

2. +w.

[2]

(ii) State, for the gas in (a), the value of

1. ΔU,

2. +q,

3. +w.

[3]

[Total: 9]

Reference: Past Exam Paper – November 2016 Paper 41 & 43 Q2

Solution:

(a)

(i)

{pV = nRT

The change occurs at constant temperature.

The amount of gas (number of moles) is also constant. R is the gas constant.

Taking into account the constant quantities,

p ∝ T

We can use proportion instead of finding each unknown quantity.}

p ∝ T               or pV / T = constant                or pV = nRT

{ p ∝ T

Initially, 1.0×10⁵ ∝ 300                        ------ (1)

After the change,   5.0×10⁵ ∝ T         ------ (2)

Divide (2) by (1),

T/300 = (5.0×10⁵)/(1.0×10⁵)

T/300 = 5}

T (= 5 × 300 =) 1500 K

(ii)

pV = nRT

{We can use either the set of initial values of the quantities or the set of final values.}

{Set of initial values:}

1.0×10⁵ × 4.0×10^-4 = n × 8.31 × 300

or

{Set of final values:}

5.0×10⁵ × 4.0×10^-4 = n × 8.31 × 1500

n = 0.016 mol

(b)

(i)

1. +q: heating/thermal energy supplied

2. +w: work done on/to system

(ii)

1. ΔU = 240 J            {as given in the question}

2. +q = 240 J   {same value as given in 1. (= 240 J) and zero given for 3.}

{ΔU = q + w

+240 = q + 0

+q = +240 J}

3. +w = zero

{Since there is no change in volume,

+w = pΔV = 0}