Question

A gas balloon has a volume of 106.0 liters when the temperature is 25.0 °C and the pressure is 740.0 mm Hg. What will its volume be at 20.0 °C and 99.3kPa

Answer 1

Answer : The final volume of gas will be, 103.3 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

$P_1$ = initial pressure of gas = 740.0 mmHg  = 98.4 kPa

Conversion used : (1 mmHg = 0.133 kPa)

$P_2$ = final pressure of gas = 99.3 kPa

$V_1$ = initial volume of gas = 106.0 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $25.0^oC=273+25.0=298K$

$T_2$ = final temperature of gas = $20.0^oC=273+20.0=293K$

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

$\frac{98.4kPa\times 106.0L}{298K}=\frac{99.3kPa\times V_2}{293K}$

$V_2=103.3L$

Therefore, the final volume of gas will be, 103.3 L