Question

0.5 mole of an ideal gas at 300 K, expands isothermally from an initial volume of 2 L to a final volume of 6 L.

Calculate:

1. work done by the gas
2. heat supplied to the gas.

[Given: R = 8.31 J mol⁻¹ K⁻¹]

Answer 1

Given: Number of moles of gas (n) = 0.5 moles

Temperature (T) = 300 K

Initial Volume (V₁) = 2 L = 2.0 × 10⁻³ m³

Final Volume (V₂) = 6 L = 6.0 × 10⁻³ m³

Universal Gas Constant (R) = 8.314 J/mol K

(a) Work done by the gas (W) = `nRT ln (V_2/V_1)`

= `0.5 xx 8.314 xx 300 xx ln(6/2)`

= `0.5 xx 8.314 xx 300 xx ln 3`

= `0.5 xx 8.314 xx 300 xx 1.0986`

= 1370 J

(b) Heat supplied to the gas (Q) = ΔU + W

For an ideal gas, the internal energy (U) is a function of temperature only. Since the process is isothermal (ΔT = 0), the change in internal energy (ΔU = 0).

∴ Q = W

= 1370 J