A cylinder of nitrogen gas has a volume of 20L at SATP. What would the volume of gas be at 30°C with a pressure of 900 kPa?

At STP: P₁ = 101 kPa, V₁ = 20 L, T₁ = 298 K
Given conditions: P₂ = 900 kPa, V₂ = ? L, T₂ = (273 + 30) K = 303 K

For a fixed amount of gas : P₁V₁/T₁ = P₂V₂/T₂
Then, V₂ = V₁ × (P₁/P₂) × (T₂/T₁)

Volume of gas at 30°C and 900 kPa = (20 L) × (101/900) × (303/298) = 2.28 L

Because of conservation of mass (and therefore conservation of the number of moles), and the formula PV = nRT (or equivalently V = nRT/P), the volume is directly proportional to the *Kelvin* temperature and inversely proportional to the pressure.

The initial temperature is (25+273.15)K = 298.15K, and the initial pressure is 1 atm = 101.325 kPa (SATP).
The final temperature is (30+273.15)K = 303.15K, and the final pressure is 900 kPa.
The initial volume is 20L.

So the final volume is 20L(303.15K / 298.15K)(101.325 kPa / 900 kPa) = 2.3L.

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