Categories

THE GAS LAWS

Pressure law

This law states that “the pressure of a fixed mass of a gas is directly proportional to the absolute temperature if the volume is kept constant”. The comparison between Kelvin scale and degrees Celsius is given by; θ0 = (273 + θ) K, and T (K) = (T – 273) 0C.

Examples

1. A gas in a fixed volume container has a pressure of 1.6 × 105 Pa at a temperature of 27 0C.

What will be the pressure of the gas if the container is heated to a temperature of 2770C?

Solution

Since law applies for Kelvin scale, convert the temperature to kelvin

T1 = 270C = (273 + 27) K = 300 K

T2 = 2270C = (273 + 277) = 550 K

P1 / T1 = P2 / T2, therefore P2 = (1.6 × 105) × 550 / 300 = 2.93 × 105 Pa.

1. At 200C, the pressure of a gas is 50 cm of mercury. At what temperature would the pressure of the gas fall to 10 cm of mercury?

Solution

P / T = constant, P1 / T1 = P2 / T2, therefore T2= (293 × 10) / 50 = 58.6 K or (– 214.4 0C)

Charles law

Charles law states that “the volume of a fixed mass of a gas is directly proportional to its absolute temperature (Kelvin) provided the pressure is kept constant”. Mathematically expressed as follows,

V1 / T1 = V2 / T2

Examples

1. A gas has a volume of 20 cm3at 270C and normal atmospheric pressure.

Calculate the new volume of the gas if it is heated to 540C at the same pressure.

Solution

Using, V1 / T1 = V2 / T2, then V2 =(20 × 327) / 300 = 21.8 cm3.

1. 0.02m3of a gas is at 27 0C is heated at a constant pressure until the volume is 0.03 m3. Calculate the final temperature of the gas in 0C.

Solution

Since V1 / T1 = V2 / T2, T2 = (300 × 0.03) / 0.02 = 450 K 0r 1770C

Boyle’s law

Boyle’s law states that “the pressure of a fixed mass of a gas is inversely proportional to its volume provided the temperature of the gas is kept constant”.

Mathematically expressed as,

P1 V1 = P2 V2

Examples

1. A gas in a cylinder occupies a volume of 465 ml when at a pressure equivalent to 725 mm of mercury. If the temperature is held constant, what will be the volume of the gas when the pressure on it is raised to 825 mm of mercury?

Solution

Using, P2 V1 = P2 V2, then V2 = (725 × 465) / 825 = 409 ml.

1. The volume of air 26 cm long is trapped by a mercury thread 5 cm long as shown below. When the tube is inverted, the air column becomes 30 cm long. What is the value of atmospheric pressure?

Solution

Before inversion, gas pressure = atm. Pressure + h p g

After inversion, gas pressure = atm. Pressure – h p g

From Boyle’s law, P1 V1 = P2 V2, then let the atm. Pressure be ‘x’,

So (x + 5) 0.26 = (x – 5) 0.30

0.26x + 1.30 = 0.3x – 1.5, x = 2.8/ 0.04 = 70 cm.

A general gas law

Any two of the three gas laws can be used derive a general gas law as follows,

P1 V1 / T1 = P2 V2 / T2 or

P V / T = constant – equation of state for an ideal gas.

Examples

1. A fixed mass of gas occupies 1.0 × 10-3m3at a pressure of 75 cmHg. What volume does the gas occupy at 17.0 0C if its pressure is 72 cm of mercury?

Solution

P V / T = constant so V= (76 × 1.0 × 10-3 × 290) / 273 ×72 = 1.12 × 10-3 m3

1. A mass of 1,200 cm3of oxygen at 270C and a pressure 1.2 atmosphere is compressed until its volume is 600 cm3and its pressure is 3.0 atmosphere. What is the Celsius temperature of the gas after compression?

Solution

Since P1 V1 / T1 = P2 V2 / T2, then T2 = (3 × 600 × 300) / 1.2 × 1,200 = 375 K or 102 0C.