Question

A monatomic ideal gas at \(27^{\circ} \mathrm{C}\) undergoes a constant volume process from \(A\) to \(B\) and a constant pressure process from \(B\) to \(C\) Find the total work done during these two processes.

Short answer

Answer: The total work done during the two processes is equal to the product of the constant pressure and the change in volume during the constant pressure process: \(W_{total} = P \Delta V\).

Step-by-step solution

Convert temperature to Kelvin

First, convert the given temperature from Celsius to Kelvin. To do this, add 273.15 to the given Celsius temperature: \(T = 27^\circ \mathrm{C} + 273.15 = 300.15\,\mathrm{K}\)

Identify the equations for work done

During a constant volume process (from A to B), the work done is given by: \(W_{AB} = 0\) (since volume is constant) During a constant pressure process (from B to C), the work done is given by: \(W_{BC} = P \Delta V\) Total work done will be the sum of work done in each process: \(W_{total} = W_{AB} + W_{BC}\)

Find the constant pressure process work done

To find the work done in the constant pressure process, we will need more information about the pressure or volume change during the process. Since this is not provided in the exercise, we cannot calculate the work done for the constant pressure process as a numerical value. However, the total work done can still be expressed in terms of pressure and volume change: \(W_{total} = 0 + P \Delta V = P \Delta V\) So, the total work done during the two processes is equal to the product of the constant pressure and the change in volume during the constant pressure process.