Question

Two liters of an ideal gas at a pressure of $10atm$ expands isothermally at ${25}^{\circ }C$ into a vacuum until its total volume is $10\text{liters}.$ How much heat is absorbed and how much work is done in the expansion? Consider the same expansion, for $1mol$ of an ideal gas conducted reversibly.

Answer 1

Step 1: Work done

As the work is done reversibly, then the work done will be given as:

$W_{rev}=-2.303nRTlog\left(\frac{V_f}{V_i}\right)$

​​ Given 𝑚𝑜𝑙𝑒 $\left(n\right)=1,T={25}^{\circ }C=298K$

$W_{rev.}=-2.303\times 1\left(mol\right)\times 0.0821\left(Latm{K}^{-1}mo{l}^{-1}\right)\times 298\left(K\right)\times log\left(\frac{10}{2}\right)$ ​​
$W_{rev.}=-2.303\times 1\times 0.0821\times 298\times 0.6990$

$W_{rev.}=-39.39Latm$

Hence, $-39.39Latm$ work is done.

Step 2: Absorbed heat

We know from first law of thermodynamics,

$△U=q+W$

As the system is working at constant temperature, i.e., isothermally.

$\therefore △U=0$

Hence, $q=-W=-(-39.39Latm)$

Therefore, $39.39Latm$ heat is absorbed.

Final answer:

$-39.39Latm$ work is done, and $39.39L$ atm heat is absorbed in the reversible expansion.