A closed gas system initially has volume and pressure of 6400mL and 1.49atm with the temperature unknown. If the same closed system has values of 657 torrs, 9660mL, and 2450C, what was the initial temperature in K?

Answer :

Answer: The initial temperature was 588 Kelvin

Explanation:

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 = 1.49 atm

[tex]P_2[/tex] = final pressure of gas = 657 torr = 0.864 atm   (760 torr = 1atm)

[tex]V_1[/tex] = initial volume of gas = 6400 ml

[tex]V_2[/tex] = final volume of gas = 9660 ml

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

[tex]T_2[/tex] = final temperature of gas = [tex]245^0C=(245+273)K=515K[/tex]

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

[tex]\frac{1.49\times 6400}{T_1}=\frac{0.864\times 9660}{515}[/tex]

[tex]T_1=588K[/tex]

The initial temperature was 588 Kelvin