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Question: One mole of a perfect gas expands isothermally and reversibly from 10 dm³ to 20 dm³ at 300 K. Find $...

One mole of a perfect gas expands isothermally and reversibly from 10 dm³ to 20 dm³ at 300 K. Find ΔU\Delta U, q and work done respectively in the process. (R = 8.3 x 10−310^{-3} kJ K−1^{-1} mol−1^{-1})

A

0.0 kJ, 1.726 kJ, -1.726 kJ

B

0.0 kJ, -17.48 kJ, -17.48 kJ

C

-1.726 kJ, 0 kJ, 1.726 k

D

0.0 kJ, 21.84 kJ, -21.84 kJ

Answer

0.0 kJ, 1.726 kJ, -1.726 kJ

Explanation

Solution

For an isothermal process of a perfect gas,
ΔU=0\Delta U = 0 (since UU depends only on TT) and from the first law, ΔU=q+w  ⟹  q=−w.\Delta U = q + w \implies q = -w.

The work done by the gas (using the sign convention where work done by the system is negative) is given by: w=−nRTln⁡VfViw = -nRT \ln\frac{V_f}{V_i}

Substitute the values: w=−1×(8.3×10−3 kJ/K mol)×300 K×ln⁡(2010)w = -1 \times (8.3 \times 10^{-3} \text{ kJ/K mol}) \times 300 \text{ K} \times \ln\left(\frac{20}{10}\right) w=−(8.3×10−3×300×ln⁡2)w = - (8.3 \times 10^{-3} \times 300 \times \ln 2) w=−(2.49×0.693)≈−1.726 kJw = - (2.49 \times 0.693) \approx -1.726 \text{ kJ}

Thus, q=−w≈+1.726 kJq = -w \approx +1.726 \text{ kJ}.

Therefore:

  • ΔU=0.0 kJ\Delta U = 0.0 \text{ kJ}
  • q=1.726 kJq = 1.726 \text{ kJ}
  • w=−1.726 kJw = -1.726 \text{ kJ}