Question

A 20.6-L sample of "pure" air is collected in Greenland at a temperature of 220.0°C and a pressure of 1.01 atm and is forced into a 1.05-L bottle for shipment to Europe for analysis.
(a) Compute the pressure inside the bottle just after it is filled.
(b) Compute the pressure inside the bottle as it is opened in the 21.0°C comfort of the European laboratory.

Answer 1

Answer :

(a) The pressure inside the bottle just after it filled is 19.8 atm.

(b) The pressure inside the bottle as it opened in the 21.0°C is 11.8 atm.

Explanation :

Part (a) :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 1.01 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 20.6 L

$V_2$ = final volume of gas = 1.05 L

Now put all the given values in the above equation, we get:

$1.01atm\times 20.6L=P_2\times 1.05L$

$P_2=19.8atm$

Therefore, the pressure inside the bottle just after it filled is 19.8 atm.

Part (b) :

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 = 19.8 atm

$P_2$ = final pressure of gas = ?

$T_1$ = initial temperature of gas = $220.0^oC=273+220.0=493.0K$

$T_2$ = final temperature of gas = $21.0^oC=273+21.0=294.0K$

Now put all the given values in the above equation, we get:

$\frac{19.8atm}{493.0K}=\frac{P_2}{294.0K}$

$P_2=11.8atm$

Therefore, the pressure inside the bottle as it opened in the 21.0°C is 11.8 atm.