91Ó°ÊÓ

• Number of moles$n=1.00\text{″¾´Ç±ô}$
• Temperature$T=0^0\text{ }C=273\text{ }K$
• Initial volume of gas$v_i=22.4\text{ L}$
• Final volume of gas$v_f=16.8\text{ L}$

In case of isothermal process temperature remains constant. The expression for the work done in case of isothermal process is given by,

$W=nRT\text{ l²Ô}\left(\frac{v_f}{v_i}\right)$

Here W is the work done,n is the number of moles, R is the gas constant,T is the temperature, $v_f$is the final volume of the gas, $v_i$is the initial volume of the gas.

The expression of the work done by an external agent is given by,

$W=nRT\text{ l²Ô}\left(\frac{v_f}{v_i}\right)$

Substitute$1.00\text{″¾´Ç±ô}$ for n, $8.31\frac{\text{J}}{\text{mol}}K$ for R, $273\text{ }K$ for T , $16.8\text{ }L$for$v_f$ , $22.4\text{ }L$for $v_i$into the above equation,

$\begin{array}{c}W=\left(1.00\text{″¾´Ç±ô}\right)\left(8.31\frac{\text{J}}{\text{mol}}K\right)\left(273\text{ }K\right)\text{ln}\left(\frac{16.8}{22.4}\right)\\ =−653\text{ J}\end{array}$

Therefore, the work done by an external agent during the isothermal compression is$−653\text{J}$ .