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Date	November 2019	Marks available	1	Reference code	19N.3.SL.TZ0.6
Level	Standard level	Paper	Paper 3	Time zone	0 - no time zone
Command term	Calculate	Question number	6	Adapted from	N/A

Question

An ideal gas consisting of 0.300 mol undergoes a process ABCD. AB is an adiabatic expansion from the initial volume V_A to the volume 1.5 V_A. BC is an isothermal compression. The pressures at C and D are the same as at A.

The following data are available.

Pressure at A = 250 kPa
Volume at C = 3.50 × 10^–3 m³
Volume at D = 2.00 × 10^–3 m³

The gas at C is further compressed to D at a constant pressure. During this compression the temperature decreases by 150 K.

For the compression CD,

Show that the pressure at B is about 130 kPa.

[2]

a(i).

Calculate the ratio $\frac{V_{A}}{V_{C}}$ .

[1]

a(ii).

determine the thermal energy removed from the system.

[3]

b(i).

explain why the entropy of the gas decreases.

[2]

b(ii).

state and explain whether the second law of thermodynamics is violated.

[2]

b(iii).

Markscheme

$P_{B} = \frac{250 \times 10^{3}}{1 . 5^{\frac{5}{3}}} « from P_{B} {(1.5 V_{A})}^{\frac{5}{3}} = 250 \times 10^{3} \times {V_{A}}^{\frac{5}{3}} »$ ✔

= 127 kPa ✔

a(i).

$« 127 \times 10^{3} \times 1.5 V_{A} = 250 \times 10^{3} V_{C} »$

1.31 ✔

a(ii).

ALTERNATIVE 1

work done $∆ W = « - » 250 \times 10^{3} \times 1.5 \times 10^{- 3} = « - » 375 « J »$ ✔

change in internal energy $∆ U = \frac{3}{2} \times 0.300 \times 8.31 \times (- 150) = « - » 561 « J »$
OR
$∆ U = \frac{3}{2} P ∆ V = \frac{3}{2} \times 375 = « - » 563 « J »$ ✔

thermal energy removed $∆ Q = 375 + 561 = 936 « J »$
OR
$∆ Q = 375 + 563 = 938 « J »$ ✔

ALTERNATIVE 2

$∆ Q = « n C p ∆ T = » \frac{5}{2} \times n R T$ ✔

thermal energy removed $∆ Q = 0.300 \times 2.5 \times 8.31 \times 150$ ✔

$= 935 « J »$ ✔

b(i).

ALTERNATIVE 1

«from b(i)» $∆ Q$ is negative ✔

$∆ S = \frac{∆ Q}{T}$ AND $∆ S$ is negative ✔

ALTERNATIVE 2

T and/or V decreases ✔

less disorder/more order «so S decreases» ✔

ALTERNATIVE 3

T decreases ✔

$∆ S = K \times \ln (\frac{T 2}{T 1}) < 0$ ✔

NOTE: Answer given, look for a valid reason that S decreases.

b(ii).

not violated ✔

the entropy of the surroundings must have increased
OR
the overall entropy of the system and the surroundings is the same or increased ✔

b(iii).

Examiners report

[N/A]

a(i).

[N/A]

a(ii).

[N/A]

b(i).

[N/A]

b(ii).

[N/A]

b(iii).

Syllabus sections

Option B: Engineering physics » Option B: Engineering physics (Core topics) » B.2 – Thermodynamics

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Option B: Engineering physics » Option B: Engineering physics (Core topics)

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