One mole of oxygen gas is at a pressure of 6.10 atm and a temperature of 25.0°C.?

If the gas is heated at constant volume until the pressure triples, what is the final temperature?

Original: P₁ = P, T₁ = (273.2 + 25.0) K = 298.2 K
New: P₂ = 3P, T₂ = ? K

For a fixed amount of gas at constant volume: P₁/T₁ = P₂/T₂
Then, T₂ = T₁ × (P₂/P₁)

New temperature, T₂ = (298.2 K) × (3P/P) = 894.6 K = 621.4 °C

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If the gas is heated so that both the pressure and volume are doubled, what is the final temperature?

Original: P₁ = P, V₁ = V, T₁ = (273.2 + 25.0) K = 298.2 K
New: P₂ = 2P, V₁ = 2V, T₂ = ? K

Combined gas law: P₁V₁/T₁ = P₂V₂/T₂
Then, T₂ = T₁ × (P₂/P₁) × (V₂/V₂)

New temperature, T₂ = (298.2 K) × (2P/P) × (2V/V) = 1192.8 K = 919.6 °C

a)
P1/T1 = P2/T2

6,10/(25+273,15) = (3*6,10)/T2

T2 = 894,45 K



b)
P1V1/T1 = P2V2/T2

6,10V1/(25+273,15) = [(2*6,10)(2V1)]/T2

T2 = 1192,60 K



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PV = nRT