the maximum work done in expending 16g of oxygen at 300k and 0ccuping a volume of 5dm3 isothermally the volume becomes 25dm3 is in5=o.7
Hello dear student,
The formula for calculating maximum work done (reversible work) is
W = -2.303*n*R*T*log(V2/V1)
RT= 8.314*300
Calculating n , given 16g of oxygen
n= (Given amount of oxygen)/(Molecular mass of Oxygen, O2)
n= 16/32 =
0.5
and, according to the question
V2 = 25 dm3
V1 = 5 dm3
Putting all the values in work done equation we get,
W = 2.303*(0.5)*8.314*300*log(25/5)
W = 2.01 x 10^3 Joule
I hope this is helpful to you,
Best Wishes!
Reply
Hello there!
Greetings!
Reverisible work is maximum work
∴w=−2.303nRTlog(V2/V1)∴w=−2.303nRTlog(V2/V1)
=2.303×16/32 ×8.314×300log25/5= 2.01×10^3 joule
thankyou
