Question

if a gas is held in 3.60 L, 298 K AND 800. mmHg, what is the new pressure of the gas if the volume us decreased by half and the tempature is decreased to 250. K

Answer 1

Apply combine gas law

`\\ \rm\Rrightarrow (P1V1)/(T1)=(P2V2)/(T2)`

`\\ \rm\Rrightarrow P2=(P1V1T2)/(V2T1)`

`\\ \rm\Rrightarrow P2=(800(3.6)(250))/(1.8(298))`

`\\ \rm\Rrightarrow P2=720000/536.4`

`\\ \rm\Rrightarrow P2=1341.7 mmHg`

Answer 2

Step-by-step explanation:

By Ideal Gas Law, P1*V1 / T1 = P2*V2 / T2

So new pressure = (P1*V1 / T1) / (V2 / T2)

= P1*V1*T2 / T1*V2

= 800*3.6*298 / 250*1.8

= 1907.2 mmHg