please I need the folowing problem in the imaige: 1.4 1.6 as soon as possible mu

Question

please I need the folowing problem in the imaige:

1.4

1.6

as soon as possible

mu A TB - mu B TA/TB - TA. where the mu and the T are the respective chemical potentials and temperatures. In a classical gas of hard spheres (of diameter sigma), the spatial distribution of the particals is no longer uncorrected. Roughly speaking, the presence of n particles in the system leaves only a volume (V - nvo) available for the (n + 1)th particle; clearly, v0 would be proportional to 1. Making use of the fact that the entropy 5(N, V, E) of a thermodynamic system is an extensive quantity, show that N(part S/part N)V.E + V(part S/part V)N.E + E(part S/part E)N.V = S. Note that this result implies: (-N mu + PV + E)/T = S, i.e. N mu = E + PV - TS. A mole of argon and a mole of helium are contained in vessles of equal volume. If argon is at 300 K, What should the temperature of helium be so that the two have the same entropy? Four moles of nitrogen and one mole of qxygon at P = 1 atm and T = 300 K are mixed together to form air at the same pressure and temperature. Calculate the entropy of mixing per mole of the air formed. Show that the various expressions for the entropy of mixing, derived in Sec. 15, satisfy the following relations: For all N1, V1, N2 and V2, (Delta S)1m2 = {(Delta S) - (Delta S)*} ge 0, the equality holding when and only when N1/V1 = N2/V2. (Delta S)* le (N1 + N2)k ln 2, the equality holding when and only when N1 =- N2. If the two gases considered in the mixing process of Sec. 1.5 were initially at different temperature, say T1 and T2, what would the entropy of mixing be in that cas? Would the contribution arising from this cause depend on whether the two gases were different or identical? Show that for an ideal gas composed of monatomic molecules the entrophy changes, between any two tempereatures, when the pressure is kept constant is 5/3 times the corresponding entrophy change when the volume is kept constant. Verify this result numerically by calculating the actual value of

Explanation / Answer

1.6)

dU = dQ + dW

Cv*dT = 10^4 + pdV = 10^4 + 0

3*R/2*(T-300) = 10^4

T = 10^4*2/(3*8.314)+300 = 1101.86 K

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