Question

${\text{5 moles}}$ of an ideal gas at ${427^0}C$ expands isothermally and reversibly from a volume of ${\text{6 L}}$ to ${\text{60 L}}$. Calculate the work done in ${\text{KJ}}$.

Answer 1

Hint: One can use the formula of the work done for isothermal reversible expansion of gas. Analyze the given values in the question and convert the values whichever is necessary.

Complete step by step answer:
1) First of all let's analyze the values given in the question and write them in their numerical denominations as follows,
Number of moles of an ideal gas ${\text{(n)}}$ ${\text{ = 5 moles}}$
Temperature $(T) = {427^0}C$
The value of temperature is given in degree Celsius but we need this value in Kelvin,
${427^0}C = 427 + 273 = 700K$
Therefore, Temperature $(T) = 700K$
Initial volume $({V_1}) = 6L$
Final volume $({V_2}) = 60L$
Work done $(W) = $ $?$
Now as we analyzed all the values given in the question now let us see the formula for the calculation of work done for the isothermal reversible process as follows,
$W = - nRT\ln \dfrac{{{V_2}}}{{{V_1}}}$
Now, let's put the values in the above formula,
$W = - 5 \times 8 \cdot 314 \times 700 \times \left( {\dfrac{{60}}{6}} \right)$
As the value of ${\text{R}}$ is common for an ideal gas.
Now let us calculate the values,
$W = - 29099 \times \left( {10} \right)$
$\Rightarrow W = - 290990{\text{ J}}$
As we need the value in kilojoules we need to convert this joules value. We can do so by dividing the joules value by one thousand to get the value in kilojoules as follows,
$W = - 290 \cdot 990{\text{ KJ}}$

Therefore, for a ${\text{5 moles}}$ of an ideal gas at ${427^0}C$ which expands isothermally and reversibly from a volume of ${\text{6 L}}$ to ${\text{60 L}}$ the work done will be ${\text{ - 290}} \cdot {\text{990 KJ}}$.


Note:
The work done value is negative which means that the work is done in the opposite direction of the force applied. The concept of isothermal expansion means the gas undergoes expansion at a constant temperature which means there is no exchange of heat between the system and surrounding. The reversible reaction means there is a constant conversion of reactant to product and product to the reactant. This process goes on until some external force or reagent is applied.