The initial temperature of three moles of oxygen gas is 32.5°C, and its pressure is 6.60 atm. (a) What will its final temperature be when heated at constant volume so the pressure is three times its initial value? °C (b) Now the volume of the gas is also allowed to change. Determine the final temperature if the gas is heated until the pressure and the volume are "quadrupled".

a. 916.95K is final temperature of the gas

b. 3667.8K

Explanation:

a. We can solve the temperature of a gas when the volume is expanded three times using Gay-Lussac's law:

P₁T₂ = P₂T₁

Where P is pressure,

And T is absolute temperature of 1, initial states and 2, final states of the gas.

Initial pressure is 6.60 atm, initial absolute temperature is 32.5°C + 273.15K = 305.65K. Final pressure is three times initial pressure = 6.60atm*3 = 19.8atm.

Solving for final temperature:

P₁T₂ = P₂T₁

6.60atm*T₂ = 19.8atm*305.65K

T₂ =

916.95K is final temperature of the gas

b.  Charle's law is the gas law that relates changes in volume and temperature of a gas:

V₁T₂ = V₂*T₁

If the volume is quadrupled: V₂/V₁ = 4:

T₂ = 4*916.95K

T₂ = 3667.8K