Question

If 0.75 mole of an ideal gas expands isothermally at \[{27^0}C\] from 15 litres to 25 litres, the maximum work obtained is:
A.-914 J
B.-956 J
C.-926 J
D.None of these

Answer 1

Hint: Isothermal is the process in which temperature is constant and gas is expanding over here so the sign of work done will be negative. Work done in this would be calculated by integrating the product of pressure and volume and using the ideal gas equation i.e. \[PV{\text{ }} = {\text{ }}nRT\] in the expression.

Complete step by step answer:
In the given question, an ideal gas is undergoing the isothermal expansion and two volumes are given, so we will use the work done formula in case of isothermal expansion of gas.
So, for an ideal gas if the gas is at initially $P_1$pressure $V_1$ volume and it reaches to $P_2$ pressure and $V_2$ pressure, the work done will be a product of Pressure and volume element \[dV\].
\[W{\text{ }} = {\text{ }}\int\limits_{V1}^{V2} {P.dV} \] if pressure is constant and volume is changing
If we integrate this and using the ideal gas equation i.e.\[PV{\text{ }} = {\text{ }}nRT\], we will finally get the expression for work that is represented as: -
\[W{\text{ }} = {\text{ }} - {\text{ }}2.303 \times {\text{ }}n \times {\text{ }}R \times {\text{ }}T{\text{ }} \times log\dfrac{{V2}}{{V1}}\]
So here we know the values of initial volume and final volume and temperature is given in question and the value of gas constant i.e. \[R{\text{ }} = {\text{ }}8.314{\text{ }}J{\text{ }}{{\text{k}}^{ - 1}}{\text{ }}mo{l^{ - 1}}\]
\[n{\text{ }} = {\text{ }}0.75{\text{ }},{\text{ }}T{\text{ }} = {\text{ }}{27^0}C{\text{ }} = {\text{ }}300K,{\text{ }}{V_2}{\text{ }} = {\text{ }}25{\text{ }}L,{\text{ }}{V_1}{\text{ }} = {\text{ }}15L\]
So, putting all these values we get
\[w = - 2.303 \times 0.75 \times 8.314 \times 298 \times \log \dfrac{{25}}{{15}}\]
$w = - 955.744 \sim 956J$
So, maximum work done by the gas is \[ - 956{\text{ }}J\]

Therefore, the correct answer is B i.e. \[ - 956{\text{ }}J\]

Note:
There are conventions in thermodynamics that the work done by the system is negative and work done on the system is positive. So here expansion is done by the system so the negative symbol is used and if compression would have been asked then the symbol of work would have positive.