211) C = Cv + kT2
dQ/ndT = dU/ndT + kT2
dQ - dU = kT2 ndT
dW = kT2ndT (1st law of thermo dynamics )
nRT dV/V = kT2ndT
R dV/V = kT dT
Now integrate
13 Answers
49Cv=Rγ-1
for a gas, γ=1+2f
so, Cv=Rf2
f=degree of freedom.!
in the given question,
n.C.dT=Q
ΔU=nCvdT=Q2
so, nCdT=2nCvdT
C=2Cv
so, C=Rf
for a monoatomic gas, f=3
Thus, cprocess=3R
21212)dQ = dU + dW
For this particular process dQ = Q
dU = Q/2 = nCvdT (because change in internal energy is independent of the path)
we know C = dQ /ndT
after putting values we get C = 2 Cv = 3R
49C = Cv + kT2
C= Cv + 1nPdVdt
so..
1nPdVdt=kT2
RTVdVdT = kT2
RdVV = kTdT
Rln(V2V1)=k2(T22-T12)
or,
RlnV-kT2/2=constant!!! :D
that is the process!
lnV=C+k2RT2
or..
V=C_{o}e^{kT^2/2R}
49we know...
Q=U+W
nCdT=nCvdT+Pdv!
tht gives the equation I used!
21little mistake in the final calculation it should be
R ln v - kt2/2 = constant
