In Problems 74 through 76 you are given the equation used to solve a problem. For each of these, you are to

a. Write a realistic problem for which this is the correct equation.

b. Finish the solution of the problem.

74.$50\mathrm{J}=-n(8.31\mathrm{J}/\mathrm{molK})(350\mathrm{K})\ln \left(\frac{1}{3}\right)$

$50\mathrm{J}=-n(8.31\mathrm{J}/\mathrm{molK})(350\mathrm{K})\ln \left(\frac{1}{3}\right)$

(a) As we can see, the formula given is the formula for the work in an isothermal process, and the work and the ratio of the initial and final volumes is given.

The only unknown is the number of moles. Therefore, we can formulate the following problem:

A gas undergoes an isothermal compression at $T=350\mathrm{K}$.

The final volume of the gas is one third of the initial volume.

In this process, the work done on the gas is $50\mathrm{J}$.

Then find the number of moles of the gas.

Realistic problem:

Find the number $n$of moles if a work of $50Joules$is done on them, with the temperature maintaining a constant value of $350\mathrm{K}$while its volume became one-third of the initial one.

$50\mathrm{J}=-n(8.31\mathrm{J}/\mathrm{molK})(350\mathrm{K})\ln \left(\frac{1}{3}\right)$

(b) The expression for the number of moles in isothermal process is

$W=-nRT\ln \left(\frac{v_f}{v_i}\right)\dots \dots$

Here,

$V_{\mathrm{f}}\to$Final volume of the gas

$V_{\mathrm{i}}\to$Initial volume of the gas

$n\to$Number of moles

$R\to$Universal gas constant$(R=8.31\mathrm{J}/\mathrm{molK})$

$T\to$Temperature of the gas

Given equation is

$(50\mathrm{J})=-n(8.31\mathrm{J}/\mathrm{molK})(350\mathrm{K})\ln \left(\frac{1}{3}\right)\dots \dots \left(2\right)$

Comparing equations (1) and (2),

The ratio of the final and initial volumes, $\frac{v_f}{v_i}=\frac{1}{3}$

Temperature of the gas, $T=350\mathrm{K}$

Work done on the gas, $W=50\mathrm{J}$

Isolate the number of moles from the equation (2)

$50\mathrm{J}=-n(8.31\mathrm{J}/\mathrm{molK})(350\mathrm{K})\ln \left(\frac{1}{3}\right)$

$n=\frac{-50\mathrm{J}}{(2908.5\mathrm{J}/\mathrm{mol})(-1.0986)}$

$=\frac{-50\mathrm{J}}{-3195.3}$

Hence, the final solution of the problem is$0.0156\mathrm{mol}$