In an isothermal process for an ideal gas, the work done by the gas can be calculated using the following formula:

$W = nRT \ln\left(\frac{V_f}{V_i}\right)$

where:

• $W$ is the work done by the gas,
• $n$ is the number of moles of the gas,
• $R$ is the ideal gas constant ($R = 8.314 \, \text{J/mol·K}$),
• $T$ is the constant temperature,
• $V_f$ is the final volume, and
• $V_i$ is the initial volume.

Since the number of moles $n$ is not given directly, we can use the ideal gas law to find it. The ideal gas law is:

$PV = nRT$

At the initial conditions ($P_i = 200 \, \text{kPa}$, $V_i = 1 \, \text{m}^3$, and $T = 300 \, \text{K}$), we can rearrange the ideal gas equation to solve for $n$:

$n = \frac{P_i V_i}{RT}$

Substituting the known values:

$n = \frac{200 \times 10^3 \, \text{Pa} \times 1 \, \text{m}^3}{8.314 \, \text{J/mol·K} \times 300 \, \text{K}}$

Now, we can compute the value of $n$, and then use that to calculate the work done during the isothermal expansion. Let's proceed with the calculation.The number of moles of the gas is approximately $80.19 \, \text{mol}$.

The work done by the gas during the isothermal expansion is approximately $219,722.46 \, \text{J}$ (or 219.72 kJ).

Would you like more details or have any further questions?

Here are 5 related questions to explore:

1. How would the work done change if the final volume was increased further?
2. How does the internal energy of an ideal gas behave during an isothermal process?
3. What is the significance of the natural logarithmic relationship in the work equation?
4. How would the process differ if it were an adiabatic expansion instead of isothermal?
5. How can you calculate the heat transfer in an isothermal process?

Tip: In an isothermal process, the temperature remains constant, meaning any work done by the gas is exactly equal to the heat absorbed.