Question

Five moles of an ideal gas expand isothermally and reversibly from an initial pressure of $100atm$ to a final pressure of $1atm$ at ${27^\circ }C$ . The work done by the gas is $(\ln 100 = 4.6)$:
(A) $2760cal$
(B) $600cal$
(C) 0
(D) $13800cal$

Answer 1

In order to answer this question, first we will rewrite the given facts and then we will convert the given temperature in its kelvin unit and then we will apply the formula of work done in terms of temperature, no. of moles and gas constant.

Complete answer:
Given that-
Number of moles of an ideal gas, $n = 5mole$
Temperature of an ideal gas that expand isothermally and reversibly, $T = {27^\circ }C = 27 + 273 = 300K$
Initial pressure of an ideal gas, ${P_1} = 100atm$
Final pressure of an ideal gas, ${P_2} = 1atm$
Now, we will apply the formula of work done in terms of temperature, moles and gas constant:-
$W = - nrt\ln (\dfrac{{{P_1}}}{{{P_2}}})$
we will put the given values of temperature, and number of moles.
as we know that, the gas contant, $r = 8.314$
and given that- $(\ln 100 = 4.6)$
$\Rightarrow W = - 5 \times 8.314 \times 300 \times \ln (\dfrac{{100}}{1})$
$\Rightarrow W = - 5 \times 8.314 \times 300 \times 4.6 \\\ \,\,\,\,\,\,\,\,\,\,\,\, = - 57366.6J \\\$
or, $W = - 13711cal \approx 13800cal$
Therefore, the required work done by the gas is $13800cal$ .
Hence, the correct option is (D) $13800cal$ .

Note:
In effect, as the gas expands, it compresses its surroundings, therefore the work done is the force exerted on the surroundings (i.e. the surrounding pressure times the area) multiplied by the distance travelled. A Joule expansion occurs when a gas expands into a vacuum, which means that the surrounding pressure is zero.