Question

A cylinder fitted with a piston contain 0.2 moles of air at temperature \[{27^ \circ }C.\]The piston is pushed so slowly that the air within the cylinder remains in thermal equilibrium with the surroundings. Find the approximate work done by the final volume is twice the initial volume
A. 549 J
B. 345 J
C. 453 J
D. 600 J

Answer 1

Hint:
 No change in temperature hence it is an isothermal process. An isothermal process is a process where the temperature of the system remains constant. You can calculate work done of an isothermal process by putting the value P means pressure from the ideal gas law equation in the equation $W = pdV$and by solving it.


Complete step by step solution:

We know that ${0^0}C = 273K$. Hence ${27^0}C = 300K$.
By looking at the question carefully, we see that water is at \[300K\] and also the ice is at \[300K\]temperature. Hence, no change in temperature. So, the process is isothermal because only change in states happens. Let’s discuss work done in an isothermal process.

Boyle's law states that the pressure and volume of a gas are inversely proportional at a constant temperature.
Now let’s solve the question.
Work done in isothermal process is given by \[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{V_2}}}{{{V_1}}}\] here ${V_1} = {\text{initial volume , }}{{\text{V}}_2} = {\text{final volume, R = universal gas constant , T = temperature,}}$
$n = {\text{number of moles}}$.
Here we are taking ${V_1} = V{\text{ and }}{{\text{V}}_2} = 2V$.
So now if we put the values as given in the question then we get
\[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{V_2}}}{{{V_1}}}\]$ = 2.306 \times 8.31 \times 0.2 \times 300 \times {\log _{10}}\left( {\dfrac{{2V}}{{1V}}} \right){\text{joule}}$$ = 345Joule(approx)$
Hence the correct option is B.



Therefore, option (B) is the correct option.



Note:
Isothermal processes happen under a constant temperature. Work done of an isothermal process is \[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{V_2}}}{{{V_1}}}\] . Alternatively, we can also write the work done in isothermal process as \[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{P_1}}}{{{P_2}}}\]. If work is done by the system, then its value will be positive and if work is done on the system then its value will be negative.