0016 mol of gas undergoes the process shown in the figure Wh

0.016 mol of gas undergoes the process shown in the figure What types of process is this? What is the final temperature in degree C? What is the final volume V_2?

Solution

a)

The volume of the gas does not remain constant so it is not isochoric

the pressure does not remain consntat so it is not isobaric

So, it must be an isothermal process <--------answer

b)

As it is an isothermal process, intital temperature = final temperature

initital temeperature, T = PV/nR = 100*10^-6*3*1.013*10^5/(0.016*8.314)

= 228.5 K <------answer

c)

final volume, V = nRT/P = 0.016*8.314*228.5/(1*1.013*10^5)

= 3*10^-4 m3

= 300 cm3