Three moles of gas initially at a pressure of 200 atm and a

Three moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 91.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 182 J.

(a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas.

(b) For the paths IAF, IBF, and IF in the figure above, calculate the net energy transferred to the gas by heat in the process.

Solution

a)

For IAF,

Wiaf = area of the curve

=(0.8-0.3)*1.013*10^5*(1.5)*10^-3

= 76 J

Similarly for IBF

Wibf = (0.8-0.3)*1.013*10^5*2*10^-3

= 101.3 J

for IF

Wif = 0.5*0.5*1.013*10^5*0.5*10^-3 + (0.8-0.3)*1.013*10^5*(1.5)*10^-3

= 88.6 J

b)

For IAF path:

For I --- > A:

W = 0

U = (3/2)nRT

= -1.5*(0.3*10^-3*0.5*1.013*10^5)

= -22.8 J

So, Q = W+ U

= U = -22.8 J

For A---- >F

W = 0.5*10^-3*1.5*1.013*10^5 = 76 J

U = 1.5*(76)

= 114 J

So, Q = W +U = 76 + 114 = 190 J

So, for the whole IAF ,

Qnet = -22.8 +190 = 167.2 J <------answer

For IBF path:

For I --- > B:

W = nRT

U = (3/2)nRT

So, Q = W+ U

= (5/2)nRT

= 2.5*2*0.5*1.013*10^5*10^-3 J

= 253.3 J

For B---- >F

W = 0

U = -1.5*(0.8*10^-3*0.5*1.013*10^5)

= -60.8 J

So, Q = W +U = -60.8 J

So, for the whole IBF ,

Qnet = 253.3 - 60.8 = 192.5 J <------answer

For IF path:

W = 0.5*0.5*10^-3*0.5*1.013*10^5 + 0.5*10^-3*1.5*10^5*1.013

= 88.6 J