Question

The volume of a gas-filled balloon is 20.0 L at 250 K and 145 kPa pressure. What would the volume be at standard temperature and pressure (STP)?

Answer 1

Answer : The volume of gas will be, 31.26 L

Solution :

At STP, pressure of gas is 101.3 kPa and temperature of gas is 273 K.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 145 kPa

$P_2$ = final pressure of gas = 101.3 kPa

$V_1$ = initial volume of gas = 20 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 250 K

$T_2$ = final temperature of gas = 273 K

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

$\frac{145kPa\times 20L}{250K}=\frac{101.3kPa\times V_2}{273K}$

$V_2=31.26L$

Therefore, the volume of gas will be, 31.26 L