At what temperature would a gas have a volume of 13.5 L at a pressure of 0.723 atm, if it had a volume of 17.8 L at a pressure of 0.612 atm and a temperature of 28 C

Answer :

Answer : The initial temperature of gas would be, 269.7 K

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 = 0.723 atm

[tex]P_2[/tex] = final pressure of gas = 0.612 atm

[tex]V_1[/tex] = initial volume of gas = 13.5 L

[tex]V_2[/tex] = final volume of gas = 17.8 L

[tex]T_1[/tex] = initial temperature of gas = ?

[tex]T_2[/tex] = final temperature of gas = [tex]28^oC=273+28=301K[/tex]

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

[tex]\frac{0.723atm\times 13.5L}{T_1}=\frac{0.612atm\times 17.8L}{301K}[/tex]

[tex]T_1=269.7K[/tex]

Thus, the initial temperature of gas would be, 269.7 K

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