Question

applying the first law of thermodynamics obtain the relation between two specific heat of a gas

Answer 1

Isometric process:

It occurs at constant V. Thus, W = 0.

From the 1st rule of thermodynamics,


∆U = Qv = n cv ∆T


Quantity of heat supplied to a system at constant V is thus equal to the increase in the internal E. of the system.


2) Isobaric process:

It occurs at constant P. e.g. water boiled and vaporized in a steam engine.

W = ∫ P dV = P ∫ dV = P (Vf – Vi)  ; Q = m Lv.

From the 1st rule of thermodynamics,


∆U = m Lv – p (Vf – Vi)

Where Vf = volume of vapor, Vi = volume of liquid.



3) Cyclic process:

In which final state and initial state are the same. Since internal E. is a state function, internal E. doesn’t change during cyclic process. Thus, ∆U = 0.

From the 1st rule of thermodynamics,


Q = W


Thus, in a cyclic process, the amount of heat gained or lost through the cycle is equal to the amount of work done = area enclosed by the cycle.


4) Isothermal process:

It occurs at constant T. e.g. in thermostats. Thus dT = 0.


P = n R T / V ---> WT = ∫ (n R T / V) dV = n R T ln (Vf / Vi).


Since T is constant, Pi Vi = Pf Vf

(Vf / Vi) = (pi / pf).


WT = n R T ln (Vf / Vi) = n R T ln (pi / pf)



5) Adiabatic process :

It occurs so that no heat flows into or out of the system. E.g. by sealing the system off from its surroundings with insulating material or by performing the process quickly (because flow of heat is slow). Thus Q = 0.

From 1st rule of thermodynamics,

∆U = Uf – Ui = – W

Answer 2

Answer:

Cp - Cv = R

Explanation:

heat supplied = 1 x Cp x dT =CvdT             ...(1)

dU = CvdT                                                     ...(2)

dQ = 1 x Cp x dT

     = CpdT                                                     ...(3)

dW = PdV

dQ = dU - dW

substituting values we get,

 CpdT = CvdT + PdV

but according to 1st law of thermodynamics

PV=nRT

PdV = RdT

=> Cp - Cv = R