# Oxygen expanded isobarically from a volume of 3 m3 to 5 m3. The oxygen pressure is 10 ^ 5 Pa.

**Oxygen expanded isobarically from a volume of 3 m3 to 5 m3. The oxygen pressure is 10 ^ 5 Pa. Determine the change in the internal energy of oxygen**

With an isobaric process p = const, then if the gas had a volume V1 at a temperature T1, then from the Mendeleev-Clapeyron equation: р ∙ V1 = ν ∙ R ∙ T1, and if then it had a volume V2 at a temperature T2, then: р ∙ V2 = ν ∙ R ∙ T2. Subtracting the first equality from the second, we get: р ∙ ΔV = ν ∙ R ∙ ΔT, where ΔТ is the temperature change, and ΔV is the change in the volume of this gas, ν is the number of moles of the taken gas and R ≈ 8.31 J / (mol ∙ K ) Is a universal gas constant. The change in the internal energy ΔU of oxygen will be ΔU = 1.5 ∙ ν ∙ R ∙ ΔТ, which means, combining the formulas, we get: ΔU = 1.5 ∙ р ∙ ΔV. It is known from the condition of the problem that oxygen has expanded isobarically from the volume V1 = 3 cubic meters. m to V2 = 5 cubic meters. m, the oxygen pressure is p = 10 ^ 5 Pa. Substitute the values of physical quantities in the calculation formula and make calculations: ΔU = 1.5 ∙ 10 ^ 5 ∙ (5 – 3); ΔU = 3 ∙ 10 ^ 5 (J).

Answer: the change in the internal energy of oxygen is 3 ∙ 10 ^ 5 J.