Question

in a sealed and rigid container, a sample of gas at 2.00 atm and 25.0 °c is heated to 50.0 °c. what is the pressure (in atm) of the gas at 50.0 °c?

Answer 1

We have that the  the pressure (in atm) of the gas at 50.0 °c is mathematically given as

P2= 2.17 atm

Gay-Lussac's law

The Gay-Lussac's law states that pressure exerted by gas is directly proportional to the absolute temperature of the gas.

Generally the equation for the Gay-Lussac's law  is mathematically given a

P1 / T1 = P2 / T2

Therefore

T1 = 25C

T1= 25 + 273

T1= 298 K

And

T2 = 50C

T2= 50 + 273

T2= 323 K

Therefore

$P2 = \frac{2.00 x 323}{298 }$

P2= 2.17 atm

Hence,  the pressure (in atm) of the gas at 50.0 °c is

$P2 = \frac{2.00 x 323}{298 }$

P2= 2.17 atm

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