An ideal gas expands isothermally at a temperature of 300 K from an initial volume of 2 L to a final volume of 5 L. If the pressure at the initial state is 4 atm, calculate the wor... An ideal gas expands isothermally at a temperature of 300 K from an initial volume of 2 L to a final volume of 5 L. If the pressure at the initial state is 4 atm, calculate the work done by the gas. Read more

Steps to Solve

1. Identify the formula for work done

In an isothermal process for an ideal gas, the formula for the work done ($W$) is given by: $$ W = nRT \ln\left(\frac{V_f}{V_i}\right) $$ where:

• $n$ = number of moles of gas
• $R$ = ideal gas constant (8.314 J/(mol·K))
• $T$ = absolute temperature in Kelvin
• $V_f$ = final volume
• $V_i$ = initial volume

2. Gather the required values

Make sure you have the values for:

• The number of moles ($n$)
• The temperature ($T$)
• The initial volume ($V_i$)
• The final volume ($V_f$)

3. Substitute the known values into the formula

Plug in the values you gathered into the formula to calculate the work done. For example, if:

• $n = 2$ moles
• $T = 300$ K
• $V_i = 1$ L (which is $0.001$ m³ for calculation)
• $V_f = 5$ L (which is $0.005$ m³ for calculation)

Then, substitute these into the formula: $$ W = 2 \cdot 8.314 \cdot 300 \cdot \ln\left(\frac{0.005}{0.001}\right) $$

4. Calculate the natural logarithm

Compute the natural logarithm: $$ \ln\left(\frac{0.005}{0.001}\right) = \ln(5) $$

5. Finish the calculation

Now multiply everything together to find the work done: $$ W = 2 \cdot 8.314 \cdot 300 \cdot \ln(5) $$ Calculate the value of $\ln(5)$, which is roughly $1.60944$, and plug it into the equation: $$ W \approx 2 \cdot 8.314 \cdot 300 \cdot 1.60944 $$

6. Find the final value for work done

Perform the multiplications to get the final answer for $W$.