Question

a gas exerts a pressure of 1.00 atm at 273 K. At what tempreture the gas exerts a pressure of 4.00 atm.

Answer 1

Considering Gay-Lussac's law, a gas exerts a pressure of 4.00 atm at 1,092 K.

Gay-Lussac's law

Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature: when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, gas pressure decreases.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T}=k$

where P= pressure, T= temperature, k= Constant

This law states that the ratio of pressure to temperature is constant.

Studying an initial state 1 and a final state 2, it is fulfilled:

$\frac{P1}{T1}=\frac{P2}{T2}$

Final temperature

In this case, you know:

• P1= 1 atm
• T1= 273 K
• P2= 4 atm
• T2= ?

Replacing in Gay-Lussac's law:

$\frac{1 atm}{273 K}=\frac{4 atm}{T2}$

Solving:

$T2\frac{1 atm}{273 K}=4 atm$

$T2=\frac{4 atm}{\frac{1 atm}{273 K}}$

T2= 1,092 K

Finally, a gas exerts a pressure of 4.00 atm at 1,092 K.

Learn more about Gay-Lussac's law:

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