Answer:
The work done is 123.5 J
Explanation:
Given that:-
Temperature = 260 K
The expression for the work done is:
[tex]W=-nRT \ln \left( \dfrac{V_2}{V_1} \right)[/tex]
Where,
n is the number of moles = 52.0 mmol = [tex]52.0\times 10^{-3}\ moles[/tex]
W is the amount of work done by the gas
R is Gas constant having value = 8.314 J / K mol
T is the temperature
V₁ is the initial volume = 300 cm³
V₂ is the final volume = 100 cm³
Applying in the equation as:
[tex]W=-52.0\times 10^{-3}\ moles\times 8.314\ J/Kmol\times 260\ K \ln \left( \dfrac{100\ cm^3}{300\ cm^3} \right)[/tex]
[tex]W=-52.0\times 10^{-3}\times 8.314\times 260 \ln \left( \dfrac{100}{300} \right)\ J=123.5\ J[/tex]
The work done is 123.5 J
