What is the final volume of a 500.0 ml gas container that increased in temperature from 299 k to 333 k while the pressure increased from 1.00 atm to 1.54 atm?

Use the combined gas law:
[tex] \frac{P _{1} V _{1} }{T _{1} } = \frac{P_{2} V _{2} }{T _{2} } [/tex]

P1 = 1.00 atm
P2 = 1.54 atm

T1 = 299 K
T2 = 333 K

T1 = 500.0 mL
T2 = ?

Solve for T2

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Alleei Alleei · Answer 2 · 2019-08-23T12:20:17+02:00

Answer : The final volume will be, 361.59 ml

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 = 1 atm

[tex]P_2[/tex] = final pressure of gas = 1.54 atm

[tex]V_1[/tex] = initial volume of gas = 500 ml

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

[tex]T_1[/tex] = initial temperature of gas = 299 K

[tex]T_2[/tex] = final temperature of gas = 333 K

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

[tex]\frac{1atm\times 500ml}{299K}=\frac{1.54atm\times V_2}{333K}[/tex]

[tex]V_2=361.59ml[/tex]

Therefore, the final volume will be, 361.59 ml