A sample of an ideal gas (3.00 l) in a closed container at 25^\circ c25∘c and 76.0 torr is heated to 300^\circ c300∘c. the pressure of the gas at this temperature is __________ torr.

So for this system, we approach ideal gas behavior for simplicity. Also we have to assume constant volume because of the hint terms 'closed container'. Let's assume the container is rigid. Then, we use the Gay-Lussac's Law.

T₁/P₁ = T₂/P₂
(25+273 K)/(76 torr) = (300+273 K)/(P₂)
Solving for P₂,
P₂ = 146.13 Torr

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Alleei Alleei · Answer 2 · 2019-12-17T11:10:15+01:00

Answer : The pressure of the gas at 300 °C is, 146 torr

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

[tex]P\propto T[/tex]

or,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 76.0 torr

[tex]P_2[/tex] = final pressure of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]300^oC=273+300=573K[/tex]

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

[tex]\frac{76.0torr}{298K}=\frac{P_2}{573K}[/tex]

[tex]P_2=146torr[/tex]

Therefore, the pressure of the gas at 300 °C is, 146 torr