## Book: HC Verma - Concepts of Physics Part 2

### Chapter: 27. Specific Heat Capacities of Gases

#### Subject: Physics - Class 12th

##### Q. No. 5 of Exercises

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5
##### The ratio of the molar heat capacities of an ideal gas is CP/CV = 7/6. Calculate the change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K(a) keeping the pressure constant,(b) keeping the volume constant and(c) adiabatically.

Given:

n = number of moles = 1,

Cv = specific heat capacity at constant volume,

Cp = specific heat capacity at constant pressure

dT = change in temperature = 50K.

γ= Ratio of molar heat capacities = CP/CV = 7/6 => Cv = 6Cp/7.

(a) Formula used:

Pressure constant: Isobaric process. For an isobaric process,

change in internal energy dU = nCvdT,

Where

n = number of moles,

Cv = specific heat at constant volume,

dT = rise in temperature

Also, Cp-Cv = R.

Cp = specific heat at constant pressure

Cv = specific heat at constant volume

R = universal gas constant = 8.314 J/mol/K

Substituting: Cp - 6Cp/7 = Cp/7 = R => Cp = 7R.

Therefore Cv = Cp - R = 6R = (6 X 8.314)J/mol/K

Therefore,

dU = 1 mol X (6 X 8.314)J/mol/K X 50K = 2494.2 J(Ans)

(b) Volume constant: Isochoric process, dV = 0(change in volume)

First law of thermodynamics gives us: dU = dQ - dW

Where dU = change in internal energy, dQ = change in heat,

dW = work done = Pressure x change in volume = PdV

Since dV = 0, dU = dQ.

Hence, dU = nCvdT since dQ = nCvdT.

Where

n = Number of moles,

Cv = Specific heat at constant volume,

dT =Change in temperature

Putting the values in the above formula, we get

Therefore, dU = 1 mol x (6 x 8.314) J/mol/K x 50K

= 2494.2 J(Ans)

(c) Adiabatic process: dQ(heat change) = 0. Therefore,

dU(change in internal energy) = dW(work done)

Since dQ = dU + dW.

For an adiabatic process, dW = dT/(

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