Work Done in Isothermal Process (using volume) Calculator

Example Calculation

Let's assume the following values:

• Number of Moles of Ideal Gas (n) = 3 Mole
• Temperature of Gas (T_g) = 300 Kelvin
• Initial Volume of System (V_i) = 11 Cubic Meter
• Final Volume of System (V_f) = 13 Cubic Meter

Using the formula:

$$W = 3 \cdot 8.31446261815324 \cdot 300 \cdot \ln\left(\frac{13}{11}\right) \approx 1250.06844792753$$

The Work Done in Isothermal Process is approximately 1250.06844792753 Joule.