$1.0 \ mol$ of a monoatomic ideal gas is expanded from state $(1)$ to state $(2)$ as shown in the figure. Calculate the work done for the expansion of the gas from state $(1)$ to state $(2)$ at $298 \ K$.

(N/A) The given diagram represents an isothermal reversible expansion of an ideal gas from pressure $P_1 = 2.0 \ bar$ to $P_2 = 1.0 \ bar$ at $T = 298 \ K$.
For an isothermal reversible process,the work done $W$ is given by the formula:
$W = -2.303 \ nRT \log_{10} \left( \frac{P_1}{P_2} \right)$
Given:
$n = 1.0 \ mol$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
$T = 298 \ K$
$P_1 = 2.0 \ bar$
$P_2 = 1.0 \ bar$
Substituting the values:
$W = -2.303 \times 1.0 \times 8.314 \times 298 \times \log_{10} \left( \frac{2.0}{1.0} \right)$
$W = -2.303 \times 8.314 \times 298 \times 0.3010$
$W \approx -1717.46 \ J \approx -1.717 \ kJ$