Question

At \( T = 27^\circ C \), \( V_i = 0.4 \, \text{L} \), \( V_f = 0.8 \, \text{L} \), find isothermal work.

Hint:

The isothermal work is calculated using the formula \( W = nRT \ln \frac{V_f}{V_i} \), where \( n \) is the number of moles and \( T \) is the temperature in Kelvin.

Answer 1

Step 1: Understanding the formula for isothermal work.
For an isothermal process, the work done by the gas is given by the formula: \[ W = nRT \ln \frac{V_f}{V_i} \] where: - \( n \) is the number of moles of the gas,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( V_f \) is the final volume, and
- \( V_i \) is the initial volume.
Step 2: Substituting the given values.
We are given: - \( T = 27^\circ C = 27 + 273 = 300 \, \text{K} \), - \( V_i = 0.4 \, \text{L} \), - \( V_f = 0.8 \, \text{L} \). Substitute these into the equation for work: \[ W = nR(300) \ln \frac{0.8}{0.4} = nR(300) \ln 2 \] Step 3: Conclusion.
Thus, the isothermal work done is \( W = nR(300) \ln 2 \). To calculate the exact value, we need the number of moles \( n \) and the value of the gas constant \( R \).

Answer 2

Step 1: For an isothermal (constant temperature) process, the work done is W = nRT × ln(V_f/V_i).

Step 2: Convert the temperature to Kelvin: T = 27°C + 273 = 300 K.

Step 3: The volume doubles from 0.4 L to 0.8 L, so V_f/V_i = 2, and ln 2 ≈ 0.693.

Step 4: So the work done is W = nR × 300 × ln 2 ≈ 208 × n × R, using the number of moles n and gas constant R.

Answer 3

Deriving the work expression from first principles.
For a gas obeying \( PV = nRT \), the pressure at any volume is \( P = nRT/V \). Work done during a volume change is \[ W = \int_{V_i}^{V_f} P \, dV = nRT \int_{V_i}^{V_f} \frac{dV}{V} = nRT \ln\frac{V_f}{V_i} \]
With \( T = 27 + 273 = 300 \, \text{K} \), \( V_i = 0.4 \, \text{L} \) and \( V_f = 0.8 \, \text{L} \), the ratio \( V_f/V_i = 2 \), so \[ W = nR(300)\ln 2 \]