Chemistry Problem Super Hard

3.0 mole of ideal gas (Cv=3R/2) at 25 °C expands reversibly and adiabatically from 10.0 atm to 1.0 atm. Calculate the work (in KJ)
a)-6.7 b)-2.2 c)-8.5 d)-12.4 e)118.6

And what's the final temperature (in K)?
a)180.3 b)196.2 c)126.5 d)232.8 e)118.6

from first law :

dU = dq + dw

since its and adiabatic expansion ...

so dq = 0

so dU = dw

and we know that for an ideal gas dU = nCvdT = nCv(T2-T1)

so nCv(Tf - Ti) = dw

now since Cv = 3R/2

and we know that Cp - Cv = R

so Cp = R + Cv = R + 3R/2 = 5R/2

now in an adiabatic process ...relation between Ti , Tf, Pi and Pf is ....
where Pi = initial pressure = 10 atm
Pf= final presure = 1 atm
Ti = initial temperature in Kelvin = 298 K
Tf = final temperature = ? K


PiTi^(-Cp/R) = PfTf^(-Cp/R)

taking log ...

log Pi + -Cp/Rlog Ti = log Pf + -Cp/R log Tf

log 10 - 5/2 log 298 = log 1 - 5/2 log Tf

1 - 5/2 X 2.474 = 0 - 5/2 log Tf

1 - 6.185 = - 2.5 log Tf

5.185 = 2.5 log Tf

log Tf = 5.185/2.5

log Tf = 2.074

taking antilog ....

Tf = 10^2.074 = 118.577 K

so final temperature is 118.6 K ...so answer for second question is e) 118.6 K

and work done = n X Cv X ( Tf - Ti)

where n = 3 mole
Cv = 3R/2
Tf = 118.6 K
Ti = 298 K
R = 8.314 J/K/mole

so W = 3 X 3/2 X 8.314 X ( 118.6-298) = 37.413 X -179.4 = -6711.892 J or -6.7 kj

so answer for first question is a) -6.7 kj

in case you dont know the relation between P and T in adiabatic process ...add additional details ..

feel free to ask any question