5.0 moles of an Ideal gas at 3.0 atm pressure and $27^{\circ} \mathrm{C}$ is compressed isothermally to half its volume by application of an external pressure of $3.5 \mathrm{~atm}$. What is the amount of work done (in joules) on the gas? Given: $1 \mathrm{~L} \mathrm{~atm}=101.3 \mathrm{~J}: \mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}{ }^{-1} \mathrm{~mol}^{-1}$

The initial state of the gas is given by $n = 5.0$ mol, $P_1 = 3.0$ atm, and $T = 27^{\circ}C = 300$ K.

Using the ideal gas law $PV = nRT$, the initial volume $V_1$ is calculated as:

$V_1 = \dfrac{nRT}{P_1} = \dfrac{5.0 \times 0.082 \times 300}{3.0} = 41$ L.

The gas is compressed isothermally to half its volume, so the final volume $V_2$ is:

$V_2 = \dfrac{V_1}{2} = \dfrac{41}{2} = 20.5$ L.

The change in volume $\Delta V = V_2 - V_1 = 20.5 - 41 = -20.5$ L.

The work done on the gas during an irreversible isothermal compression against a constant external pressure $P_{ext}$ is given by $W = -P_{ext} \Delta V$.

Given $P_{ext} = 3.5$ atm, the work done is:

$W = -3.5 \times (-20.5) = 71.75$ L atm.

Converting the work into Joules using the conversion factor $1$ L atm $= 101.3$ J:

$W = 71.75 \times 101.3 = 7268.275$ J.

Rounding to one decimal place, the work done is $7268.3$ J.

Answer: 7268.3