Thermodynamics

To calculate the work done by the gas in an isothermal process, we proceed as follows:

Use the formula for work done in an isothermal process: $W%20%3D%20nRT%20%5Cln%5Cleft(%5Cfrac%7BV_f%7D%7BV_i%7D%5Cright).$
Given: $n%20%3D%201%20%5C%2C%20%5Ctext%7Bmole%7D$ , $R%20%3D%208.314%20%5C%2C%20%5Ctext%7BJ%2FmolK%7D$ , $T%20%3D%20300%20%5C%2C%20%5Ctext%7BK%7D$ , $V_i%20%3D%202%20%5C%2C%20%5Ctext%7BL%7D$ , $V_f%20%3D%204%20%5C%2C%20%5Ctext%7BL%7D$ .
Calculate the volume ratio: $%5Cfrac%7BV_f%7D%7BV_i%7D%20%3D%20%5Cfrac%7B4%7D%7B2%7D%20%3D%202.$
Find the natural logarithm: $%5Cln(2)%20%5Capprox%200.693.$
Compute $nRT$ : $nRT%20%3D%201%20%5Ctimes%208.314%20%5Ctimes%20300%20%3D%202494.2%20%5C%2C%20%5Ctext%7BJ%7D.$
Calculate work done: $W%20%3D%202494.2%20%5Ctimes%200.693%20%5Capprox%201728.5%20%5Ctimes%200.482%20%5Capprox%20831.4%20%5C%2C%20%5Ctext%7BJ%7D.$
Match with options: 831.4 J corresponds to option (B).

Thus, the correct answer is: $%5Cboxed%7B831.4%20%5C%2C%20%5Ctext%7BJ%7D%7D$

Chapter Questions
2 MCQs