Question

What will be the work done when one mole of gas expands isothermally from 15L to 50L against a constant pressure of 1 atm at $25^oC$.
A.-3542cal
B.-843.3cal
C.-718cal
D.-60.23cal

Answer 1

Isothermal expansion of gas is defined as the increase in the volume of gas under constant temperature. During the isothermal expansion, work is done by the gas so work done obtained is negatively signed.

Complete step by step answer:

The work done for an isothermal expansion is given by the following formula:
${\text{Work done(W) = - nRTln}}(\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{V}}_{\text{1}}}}})$ or
${\text{Work done(W) = 2}}{\text{.303}} \times {\text{ - nRTlog}}(\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{V}}_{\text{1}}}}})$
Where Work done by the system is represented by W, n represents number of moles, R is gas constant and T is temperature. Also, V1 and V2 are initial and final pressure respectively.
As per numerical above, the given values for different parameters is shown below:

& {{V}_{1}}=15L \\\ & {{V}_{2}}=50L \\\ & R=8.314J/K/mol \\\ & n=1\text{ }mol \\\ & T={{25}^{{}^\circ }}C\text{ }or\text{ }\left( 25+273=298K \right) \\\ \end{aligned}$$ On putting the value in the above formula we get: ${\text{Work done(W) = - 2}}{\text{.303}} \times 1 \times 8.314 \times 298 \times {\text{log}}(\dfrac{{50}}{{15}})$ it can also be written as: ${\text{Work done(W) = - 2}}{\text{.303}} \times 1 \times 8.314 \times 298 \times {\text{log}}(\dfrac{{10}}{3})$ (1) And as we know $\log \dfrac{{\text{A}}}{{\text{B}}} = {\text{logA - logB}}$. Hence, the above equation becomes: ${\text{Work done(W) = - 2}}{\text{.303}} \times 1 \times 8.314 \times 298 \times (\log 10 - \log 3)$ Since the value $\log 10$is 1 and the value of $\log 3$ is 0.477 on solving we get: ${\text{Work done(W) = - 2}}{\text{.303}} \times 1 \times 8.314 \times 298 \times (1 - 0.477)$ $\Rightarrow \text{Work done(W) = - 2}\text{.303}\times 1\times 8.314\times 298\times 0.523$ $\Rightarrow \text{Work done(W) = - 717}\text{.86 cal }\approx \text{ - 718 cal}$ Hence the work done by one mole of gas during isothermal expansion is -718 cal **Hence, option C is the correct answer.** **Note:** Isothermal process is an idealized process and hence it simplifies calculations and it simplifies the use of the first law of thermodynamics. We assume the temperature to be constant in this process and hence Boyle’s law and ideal gas law can be used to relate pressure and volume. Since, all the heat energy is converted to work without any loss, this process is considered to be a reversible process.