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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 99.3 kPa?

Answer :

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,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 740.0 mmHg = 98.4 kPa

Conversion used : (1 mmHg = 0.133 kPa)

[tex]P_2[/tex] = final pressure of gas = 99.3 kPa

[tex]V_1[/tex] = initial volume of gas = 106.0 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]25.0^oC=273+25.0=298K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]20.0^oC=273+20.0=293K[/tex]

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

[tex]\frac{98.4kPa\times 106.0L}{298K}=\frac{99.3kPa\times V_2}{293K}[/tex]

[tex]V_2=103.3L[/tex]

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

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