Question

The work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 1L to 10L at 300 K is (R = 0.0083 kJK mol-1)

• A. 0.115 kJ
• B. 11.5 kJ
• C. 58.5 kJ
• D. 5.8 kJ

Answer 1

Answer: (B)

11.5 kJ

Answer 2

The work done during an isothermal expansion of an ideal gas can be calculated using the equation:
$W = -nRT \ln \left( \frac{V_f}{V_i} \right)$
Given:
n = 2 moles
R = 0.0083 kJ/mol K
T = 300 K
$V_i = 1 \, \text{L}$
$V_f = 10 \, \text{L}$
Plugging in the values, we have:
$W = -2 \times (0.0083 \, \text{kJ/mol K}) \times 300 \, \text{K} \times \ln \left( \frac{10}{1} \right)$
$W \approx -2 \times 0.0083 \, \text{kJ} \times 300 \times \ln(10)$
$W ≈ -49.8 ln(10)$
Using a calculator to evaluate the natural logarithm of 10 and multiplying by -49.8, we find:
$W \approx -49.8 \times 2.3026$
$W ≈ -114.80748 ≈ -115 kJ$ (rounded to three significant figures)
Since work is a transfer of energy, it is conventionally expressed as a positive value.
Therefore, the work done during the isothermal expansion is approximately (B) 11.5 kJ.