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,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

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.

$\frac{29.4kPa}{348K}=\frac{P_2}{298K}$

$P_2=25.176kPa$

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