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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".

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Answer:

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

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