Answer :

Given:


P = kV …


Where,


P = pressure,


V = volume,


k = constant.


Formula used:


Equation of state of ideal gas:


PV = nRT = constant … (ii)


Where,


n is the number of moles of the gas,


R is the gas constant,


T is the temperature,


P = pressure, V2


T = temperature.


From (i), multiplying by dV on both sides:


PdV = kVdV.


Integrating from V = V1 to V2, we get



= with lower limit V1 and upper limit V2


=


Now, we know, PV = nRT - equation of state,


Where P = pressure, V = volume, n = number of moles, R = universal gas constant, T = temperature


Hence we can write, V1 = nRT1/P1. Since P1 = kV1, this becomes:


kV12 = nRT1. Similarly, KV22 = nRT2,


P1, V1, T1 - Pressure, volume, temperature of first gas


P2, V2, T2 - Pressure, volume, temperature of second gas


Therefore, subsituting, the above equation becomes:


= = … (iii)


Now, => (since P = kV) … (iv)


But Q = U + PdV (first law of thermodynamics), where Q = heat, U = change in internal energy, W = total work done = PdV


=> nCdT = nCvdT + (nR/2)dT


(since Q = nCdT and U = nCvdT)


=> C = Cv + nR/2 (proved),


Where


n = number of moles,


C = specific heat capacity,


Cv = specific heat capacity at constant volume,


R = universal gas constant,


dT = rise in temperature.


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