##### An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p = kV. Show that the molar heat capacity of the gas for the process is given by .

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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