Question

calculate the total work done when one mole of gas expands isothermally and reversibly from an initial value of 20 DM cube to A final volume of 40 DM cube at 298 k ( R=8.314 J

Answer 1

Answer:

The total work done when one mole of gas expands isothermally and reversibly from an initial value of $20\ dm^3$ to a final volume of $40\ dm^3$ at $298$ K  is $-\ 1717.460$ K J·

Explanation:

Given that,

The number of moles of gas $=$ $1$ mol

The initial volume $=$ $20\ dm^3$

The final volume $=$ $40\ dm^3$

Temperature $=$ $298$ K

The value of R $=$ $8.314\ J\ K^{-1}\ mol^{-1}$

It is mentioned that the gas expands isothermally and reversibly·

In order to find out the work done,

we know that the equation to calculate the work done in an isothermal reversible expansion process of an ideal gas follows,

    W $=\ -\ 2.303$ n R T log $\frac{V_f}{V_i}$

where,

n $=$ number of moles of a gas

R $=$ Universal gas constant

T $=$ Temperature in kelvin

$V_i$ $=$ The initial volume

$V_f$ $=$ The final volume

On substituting the values in the equation, we get

 ⇒  W $=\ -\ 2.303$ n R T log $\frac{V_f}{V_i}$

 ⇒  W  $=$ $-2.303$ × $1$ × $8.314$ × $298$ × log $\frac{40}{20}$

 ⇒  W $=$  $-\ 5705.848$ × $0.301$

 ⇒  W  $=$ $-\ 1717.460$ K J

Therefore,

The total work done when one mole of gas expands isothermally and reversibly is  $-\ 1717.460$ K J·