Question

A helium gas tank at 3.00 x 103 mm Hg is cooled from 500 Kelvin to 300 Kelvin. What is the final pressure of the helium gas in the steel tank?

Answer 1

`P_2=1.80x10^3mmHg`

Step-by-step explanation:

Hello,

In this case, we use the Gay-Lussac's law to understand the pressure-temperature relationship a directly proportional relationship:

`(P_1)/(T_1) =(P_2)/(T_2)`

Thus, solving for the final pressure P2, we obtain:

`P_2=(P_1T_2)/(T_1)=(3.00x10^3mmHg*300K)/(500K)\\ \\P_2=1.80x10^3mmHg`

Best regards.