User interface language: English | Español

Date May 2018 Marks available 2 Reference code 18M.3.SL.TZ2.7
Level Standard level Paper Paper 3 Time zone Time zone 2
Command term Show that Question number 7 Adapted from N/A

Question

A cylinder is fitted with a piston. A fixed mass of an ideal gas fills the space above the piston.

M18/4/PHYSI/SP3/ENG/TZ2/07._01

The gas expands isobarically. The following data are available.

\[\begin{array}{*{20}{l}} {{\text{Amount of gas}}}&{ = 243{\text{ mol}}} \\ {{\text{Initial volume of gas}}}&{ = 47.1{\text{ }}{{\text{m}}^3}} \\ {{\text{Initial temperature of gas}}}&{ = -12.0{\text{ °C}}} \\ {{\text{Final temperature of gas}}}&{ = + 19.0{\text{ °C}}} \\ {{\text{Initial pressure of gas}}}&{ = 11.2{\text{ kPa}}} \end{array}\]

The gas returns to its original state by an adiabatic compression followed by cooling at constant volume.

Show that the final volume of the gas is about 53 m3.

[2]
a.

Calculate, in J, the work done by the gas during this expansion.

[2]
b.

Determine the thermal energy which enters the gas during this expansion.

[3]
c.

Sketch, on the pV diagram, the complete cycle of changes for the gas, labelling the changes clearly. The expansion shown in (a) and (b) is drawn for you.

[2]
d.i.

Outline the change in entropy of the gas during the cooling at constant volume.

[1]
d.ii.

There are various equivalent versions of the second law of thermodynamics. Outline the benefit gained by having alternative forms of a law.

[1]
e.

Markscheme

ALTERNATIVE 1

«Using \(\frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}}\)»

V2 = \(\frac{{47.1 \times (273 + 19)}}{{(273 - 12)}}\)

V2 = 52.7 «m3»

 

ALTERNATIVE 2

«Using PV = nRT»

V = \(\frac{{243 \times 8.31 \times (273 + 19)}}{{11.2 \times {{10}^3}}}\)

V = 52.6 «m3»

 

[2 marks]

a.

W «PΔV» = 11.2 × 103 × (52.7 – 47.1)

W = 62.7 × 103 «J»

 

Accept 66.1 × 103 J if 53 used

Accept 61.6 × 103 J if 52.6 used

[2 marks]

b.

ΔU «= \(\frac{3}{2}\)nRΔT» = 1.5 × 243 × 8.31 × (19 – (–12)) = 9.39 × 104

Q «= ΔUW» = 9.39 × 104 + 6.27 × 104

Q = 1.57 × 105 «J»

 

Accept 1.60 × 105 if 66.1 × 103 J used

Accept 1.55 × 105 if 61.6 × 103 J used

[3 marks]

c.

concave curve from RHS of present line to point above LHS of present line

vertical line from previous curve to the beginning

 

M18/4/PHYSI/SP3/ENG/TZ2/07.d.i/M

 

[2 marks]

d.i.

energy is removed from the gas and so entropy decreases

OR

temperature decreases «at constant volume (less disorder)» so entropy decreases

 

OWTTE

[1 mark]

d.ii.

different paradigms/ways of thinking/modelling/views

allows testing in different ways

laws can be applied different situations

 

OWTTE

[1 mark]

e.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
e.

Syllabus sections

Option B: Engineering physics » Option B: Engineering physics (Core topics) » B.2 – Thermodynamics
Show 114 related questions

View options