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How to solve questions on gas law for UTME

 

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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

 
short notes on gas law for UTME

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