chapter  31
9 Pages

## Ideal gas laws

Problem 1. A gas occupies a volume of 0.10 m3 at a pressure of 1.8 MPa. Determine (a) the pressure if the volume is changed to 0.06 m3at constant temperature and (b) the volume if the pressure is changed to 2.4 MPa at constant temperature

(a) Since the change occurs at constant temperature (i.e. an isothermal change), Boyle’s law applies, i.e. p1V1 = p2V2 where p1 = 1.8 MPa, V1 = 0.10 m3 and V2= 0.06 m3 Hence, (1.8)(0.10) = p2(0.06) from which,

pressure p2 = 1.8 × 0.10

0.06 = 3 MPa

(b) p1 V1 = p2 V2 where p1 = 1.8 MPa, V1 = 0.10 m3 and p2 = 2.4 MPa. Hence, (1.8)(0.10) = (2.4)V2 from which,

volume V2 = 1.8 × 0.10

2.4 = 0.075 m3

Problem 2. In an isothermal process, a mass of gas has its volume reduced from 3200 mm3 to 2000 mm3. If the initial pressure of the gas is 110 kPa, determine the final pressure

Since the process is isothermal, it takes place at constant temperature and hence Boyle’s law applies, i.e. p1V1 = p2V2, where p1 = 110 kPa, V1 = 3200 mm3 and V2 = 2000 mm3. Hence, (110)(3200) = p2(2000) from which,

final pressure p2 = 110 × 3200

2000 = 176 kPa

Problem 3. Some gas occupies a volume of 1.5 m3 in a cylinder at a pressure of 250 kPa. A piston, sliding in the cylinder, compresses the gas isothermally until the volume is 0.5 m3. If the area of the piston is 300 cm2, calculate the force on the piston when the gas is compressed

An isothermal process means constant temperature and thus Boyle’s law applies, i.e. p1V1 = p2V2, where V1 = 1.5 m3, V2 = 0.5 m3 and p1 = 250 kPa. Hence, (250)(1.5) = p2 (0.5) from which,

pressure p2 = 250 × 1.5

0.5 = 750 kPa

Pressure = force area

, from which,

force = pressure × area. Hence, force on the piston

= (750 × 103Pa) (300 × 10−4m2) = 22.5 kN

Now try the following Practice Exercise

1. The pressure of a mass of gas is increased from 150 kPa to 750 kPa at constant temperature. Determine the final volume of the gas, if its initial volume is 1.5 m3

2. In an isothermal process, a mass of gas has its volume reduced from 50 cm3 to 32 cm3. If the initial pressure of the gas is 80 kPa, determine its final pressure

3. The piston air to

1 4

of stroke. if the original pressure is 100 kPa, assuming an isothermal change

4. A quantity of gas in a cylinder occupies a volume of 2 m3 at a pressure of 300 kPa. A piston slides in the cylinder and compresses the gas, according to Boyle’s law, until the volume is 0.5 m3. If the area of the piston is 0.02 m2, calculate the force on the piston when the gas is compressed

Charles’ law states:

For a given mass of gas at constant pressure, the volume V is directly proportional to its thermodynamic temperature T

i.e. V ∝ T or V = kT or V T

= k at constant pressure, where T = thermodynamic temperature in kelvin (K). A process that takes place at constant pressure is called an isobaric process. The relationship between the Celsius scale of temperature and the thermodynamic or absolute scale is given by:

kelvin = degrees Celsius + 273 i.e. K = ◦C + 273 or ◦C = K − 273

(as stated in chapter 29). If a given mass of gas at a constant pressure occupies a volume V1 at a temperature T1 and a volume V2 at temperature T2, then

V1 T1

= V2 T2

Problem 4. A gas occupies a volume of 1.2 litres at 20◦C. Determine the volume it occupies at 130◦C if the pressure is kept constant

(i.e. an isobaric process), Charles’ law applies,

i.e. V1 T1

= V2 T2

where V1 = 1.2 litre, T1 = 20◦C = (20 + 273)K = 293 K and T2 = (130 + 273)K = 403 K.