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ThermodynamicsEasy

Ideal gas — isobaric process

An ideal gas occupies V₁ = 2 L at T₁ = 300 K at constant pressure.\nWe heat it to T₂ = 450 K. Calculate the new volume V₂ and the work done.
Given data
V₁ = 2 L = 2×10⁻³ m³T₁ = 300 KT₂ = 450 KP = constant
Review the theory: Termodinamica
Steps
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  1. What is the volume V₂ of the gas after heating to T₂ = 450 K? (in litres)
  2. How much work does the gas do while expanding? (P = 10⁵ Pa, ΔV in m³)
Full worked solution
  1. What is the volume V₂ of the gas after heating to T₂ = 450 K? (in litres)
    V2=V1T2T1=2⋅450300V_2 = V_1 \dfrac{T_2}{T_1} = 2 \cdot \dfrac{450}{300}
    V2=V1⋅T2/T1=2×450/300=2×1.5=3 LV_2 = V_1 \cdot T_2/T_1 = 2 \times 450/300 = 2 \times 1.5 = \mathbf{3\,L}. At constant pressure, volume is proportional to absolute temperature (Charles's law V/T=constV/T = \text{const}).
  2. How much work does the gas do while expanding? (P = 10⁵ Pa, ΔV in m³)
    L=P⋅ΔV=105⋅(3−2)×10−3L = P \cdot \Delta V = 10^5 \cdot (3 - 2) \times 10^{-3}
    W=P ΔV=105×(3×10−3−2×10−3)=105×10−3=100 JW = P\,\Delta V = 10^5 \times (3\times10^{-3} - 2\times10^{-3}) = 10^5 \times 10^{-3} = \mathbf{100\,J}. In an isobaric expansion, the work done by the gas is pressure times the change in volume.
Result:V₂ = 3 L, Work done = 100 J (at P = 10⁵ Pa).
TheoryTermodinamicaFormula sheetPhysics 1

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