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Worked Examples · Example 4.1

Q.0.5 mole of gas at temperature 300 K expands isothermally from an initial volume of 2.0 L to a final volume of 6.0 L.

(a) What is the work done by the gas? (R = 8.319 J mol⁻¹ K⁻¹)
(b) How much heat is supplied to the gas?
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W = 0.5 × 8.319 × 300 × ln(6/2) = 1.369 kJ; since ΔU = 0, Q = W = 1.369 kJ.

(a) Work done. For an isothermal expansion,

W=nRTln⁡ ⁣(VfVi)=0.5×8.319×300×ln⁡ ⁣(6.02.0)=1.369 kJW = nRT\ln\!\left(\frac{V_f}{V_i}\right) = 0.5 \times 8.319 \times 300 \times \ln\!\left(\frac{6.0}{2.0}\right) = 1.369\ \mathrm{kJ} …

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