Question

A gas at 29.4 kPa is cooled from a temperature of 75°C to a temperature of 25°C at constant volume. What is the new pressure of the gas?

Answer 1

Answer : The new pressure of gas will be, 25.176 kPa

Solution :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

`P\propto T`

or,

`(P_1)/(T_1)=(P_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 29.4 kPa

`P_2` = final pressure of gas = ?

`T_1` = initial temperature of gas =
`75^oC=273+75=348K`

`T_2` = final temperature of gas =
`25^oC=273+25=298K`

Now put all the given values in the above equation, we get the final pressure of gas.

`(29.4kPa)/(348K)=(P_2)/(298K)`

`P_2=25.176kPa`

Therefore, the new pressure of gas will be, 25.176 kPa

Answer 2

To solve this we assume that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant volume pressure and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

T1/P1 = T2/P2

P2 = T2 x P1 / T1

P2 = 25 x 29.4 / 75

P2 = 9.8 kPa