How to solve questions on gas law for UTME
Question 1
The pressure of a given mass of a gas changes from 300Nm-2 to 120Nm-2 while the temperature drops from 127oC to –73oC. The ratio of the final volume to the initial volume is{UTME 2001}
A. 2 : 5 B. 4 : 5 C. 5 : 2 D. 5 : 4
Solution
P1V1/T1 = P2V2/T2
127 degree = 273+127 = 400 k
-73 degree = 273-73 = 200 k
300*V1/400 = 120*V2/200
300*200*V1 = 120*400*V2
V1/V2 = 120*400/300*200 = 24/30 = 4:5
V2/V1 = 5:4
D is the correct option
Question 2
The pressure of 3 moles of an ideal gas at a temperature of 27oC having a volume of 10-3m3 is{UTME 2002}
A. 2.49 x 105Nm-2 B. 7.47 x 105Nm-2 C. 2.49 x 106Nm-2 D. 7.47 x 106Nm-2
[R = 8.3J mol-1K-1]
Solution
PV = nRT
P = nRT/V
P = 3*8.3*(273+27)/10-3
P = 3*8.3*300/0.001
P = 7470000Nm-2 = 7.47 x 106Nm-2
D is correct option
Question 3
Which of the following gas laws is equivalent to the work done?{UTME 2007}
A. Pressure Law B. Van der Waal’s Law C. Boyle’s Law D. Charles’ Law
Solution
PV = K(k=constant), this is Boyle’s Law and is equivalent to workdone
C is the correct option
Question 4
A sealed flask contains 600cm3 of air at 27oC and is heated to 35oC at constant pressure. The new volume is{UTME 2008}
A. 508cm3 B. 516cm3 C. 608cm3 D. 616cm3
Solution
V1/T1 = V2/T2
600/(273+27) = V2/(273+35)
600/300 = V2/308
2/1 = V2/308
V2 = 308*2 = 616cm3
D is the correct option
Question 5
At 40C, the volume of a fixed mass of water is{UTME 2009}
A. constant B. minimum C. maximum D. zero.
Solution
Anomalous behaviour of water is between 0 – 4 degree. At this temperature water contract i.e volume of water decreases. So the volume will be minimum
B is the correct option
Question 6
The pressure of two moles of an ideal gas at a temperature of 270C and volume 10-2m3 is{UTME 2009}
A. 4.99 x 105 Nm-2 B. 9.80 x 103 Nm-2 C. 4.98 x 103 Nm-2 D. 9.80 x 105 Nm-2
[R = 8.313 J mol-1 K-1]
Solution
PV = nRT
P = nRT/V
P = 2*8.313*(273+27)/10-2
P = 2*8.313*300/0.01
P = 498780Nm-2 = 4.99 x 105 Nm-2
A is correct option
Question 7
The pressure of one mole of an ideal gas of volume 10-2m3 at a temperature of 270C is {UTME 2010}
A. 2.24 x 104 Nm-2 B. 2.24 x 105 Nm-2 C. 2.49 x 105 Nm-2 D. 2.49 x 104 Nm-2.
[Molar gas constant = 8.3 Jmol-1K-1]
Solution
PV = nRT
P = nRT/V
P = 1*8.3*(273+27)/10-2
P = 1*8.313*300/0.01
P = 249000Nm-2 = 2.49 x 105 Nm-2
C is correct option
Question 8
2000cm3 of a gas is collected at 27oC and 700mmHg. What is the volume of the gas at standard temperature and
pressure?{UTME 2012}
A. 1896.5cm3 B. 1767.3cm3 C. 1676.3cm3 D. 1456.5cm3
Solution
At s.t.p pressure = 760mmHg, temperature = 273K
P1V1/T1 = P2V2/T2
700*200/300 = 760*V2/273
V2 = 700*2000*273/(300*760) =1676.3cm3
C is the correct option
Question 9
A gas at a pressure of 105Nm-2 expands from 0.6m3 to 1.2m3 at constant temperature, the work done is{UTME 2013}
A. 6.0 x 104J B. 7.0 x 107J C. 6.0 x 106J D. 6.0 x 105J
Solution
workdone = P(V2 – V1)
workdone = 100000(1.2 – 0.6) = 100000(0.6)
workdone = 60000 J = 6.0 x 104J
A is the correct option
