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,

`(P_1V_1)/(T_1)=(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:

`(98.4kPa* 106.0L)/(298K)=(99.3kPa* V_2)/(293K)`

`V_2=103.3L`

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