# Exam-style Questions: Thermodynamics and Ideal Gases

1. Study the graphs A, B, C, D

a) Which graph shows the variation in volume (y) of a fixed mass of ideal gas at constant pressure with absolute temperature (x)?

(1 Mark)

b) Which graph shows the variation in pressure (y) of a fixed mass of ideal gas at constant temperature with volume (x)?

(1 Mark)

(Marks available: 2)

Answer outline and marking scheme for question: 1

(1 Mark)

(1 Mark)

(Marks available: 2)

2. 4.0 g of helium contains one mole (6.0 x 1023 atoms). The helium is at a presure of 1.0 x 105 Pa and at a temperature of 300 K.

a) Show that one mole of helium occupies a volume of about 0.025 m3 under these conditions.

molar gad constant R = 8.3 J mol-1 K-1

(2 Marks)

b) The gas is compressed to volume of 0.020 m3. The temperature of the gas is kept constant.

Calculate the new pressure of the gas.

pressure = ..................... pA

(2 Marks)

(Marks available: 4)

a) pV = nRT (1 Mark) -> V = nRT/p = 1 x 8.31 x 300 / 1.0 x 105 (1 Mark) = 0.0249 m3

(can work backwards or use pV = nKt)

(2 Marks)

b) 1.0 X 105 X 0.025 = P2 X 0.020 (1 Mark) p2 = 1.3 x 105 Pa (1 Mark) (accept 1.25 x 105)

candidates can use their answers to 5 a) (ecf) (e.g. 1.2 x 105 if 0.0249 used).

(2 Marks)

(Marks available: 4)

3. Two students attempt the same experiment to find how air pressure varies with temperature. They heat identical sealed glass flasks of air, to be considered as an ideal gas, in an oil bath. The flasks are heated from 300 K to 400 K. The pressure in flask A rises from atmospheric presure, po, as expected, but the pressure in flask B remains at po because the rubber bung is defective and air leaks out of the flask.

a) Calculate the pressure in flask A at 400 K in terms of po.

pressure = ........................

(2 Marks)

b) Calculate the fraction, f, of gas molecules in flask B compared to flask A at 400 K.

 f = number of gas molecule in B at 400 K number of gas molecule in A at 400 K

f = ...............................

(2 Marks)

(Marks available: 4)

a) p/T = constant / AW; p/po = T1/To = 400/300 giving p = 1.33 po

(2 Marks)

b) use of n = pV/RT / n (proportional to) p / N (proportional to) p / f = (NB/NA) = nB/nA = po/p

= ¾ or 0.75 ecf from c(i)

(2 Marks)

(Marks available: 4)