Question

Calculate the maximum work done (in J) in expanding 16g of oxygen at300K and occupying a volume of 5$d{{m}^{3}}$isothermally until the volume becomes 25 $d{{m}^{3}}$
(A) $+2.11\text{ X }{{10}^{3}}$
(B) $+2.01\text{ X }{{10}^{3}}$
(C) $-2.01\text{ X }{{10}^{3}}$
(D) $-2.11\text{ X }{{10}^{3}}$

Answer 1

In the isothermal process is a change in the system in which the temperature of the system remains constant. An example of isothermal process is change in state or phase changes in the different liquid by the process of melting and evaporation.

Complete step by step solution:
Given in the question:
Temperature = 300K
Given mass of oxygen = 16 g
Volume ${{V}_{1}}$ (initial) = 5$d{{m}^{3}}$
Volume ${{V}_{2}}$ (final) = 25$d{{m}^{3}}$
The formula use for the calculation of maximum work done
W= -2.303nRTlog$\dfrac{{{V}_{2}}}{{{V}_{1}}}$
Where,
W = maximum work done
R = universal gas constant
T = temperature
And ${{V}_{1}}$ and ${{V}_{2}}$ is the initial and final volume
Value of R i.e. universal gas constant = 8.314$Jmo{{l}^{-1}}{{K}^{-1}}$
Calculation of number of moles of oxygen
Molar mass of oxygen= 32
Number of moles is calculated by the ratio of given mass by molar mass.
Number of moles = $\dfrac{given\text{ }mass}{molar\text{ mass}}$
Number of moles of oxygen = 0.5 mole
Now put all the value i.e. number of moles, ${{V}_{1}}$ and ${{V}_{2}}$ in the equation of work done
W= -2.303nRTlog$\dfrac{{{V}_{2}}}{{{V}_{1}}}$
W= $-2.303\left( 0.5 \right)\left( 8.314 \right)\left( 300 \right)\log \dfrac{25}{5}$
After solving the equation
W= -2.01$X{{10}^{3}}$joule

Hence the correct option is option (C)

The maximum work done (in J) in expanding 16g of oxygen at300K and occupying a volume of 5$d{{m}^{3}}$isothermally until the volume becomes 25 $d{{m}^{3}}$ is -2.01$X{{10}^{3}}$joule.

Note: Logarithm in this question is solved using the formula of quotient rule .Which means that the logarithm of quotient of two factors to any base which is positive but not 1 is equal to the difference of the logarithm of the factor to the same base.