A sample of air weighing 1.18g occupies 1.0 × 103 cm3 when kept at 300K and 1.0 × 105 Pa. When 2.0 cal of heat is added to it at constant volume, its temperature increases by 1°C. Calculate the amount of heat needed to increase the temperature of air by 1°C at constant pressure if the mechanical equivalent of heat is 4.2 × 107 erg cal–1. Assume that air behaves as an ideal gas.

Given:


Pressure, p= 1.0 × 105 Pa,


Temperature, T = 300K,


Universal gas constant, R = 8.314 J/kg/mol


V(volume) = 1.0 × 103 cm3 = 0.001 m3


Formula used:


1. Ideal gas equation: PV = nRT.


Where,


P = pressure,


V = volume,


n = number of moles,


R = universal gas constant = 8.314 J/kg/mol,


T = absolute temperature


2. Number of moles n = PV/RT = 100000x0.001/(8.314*300) = 0.04 mol


3. First law of thermodynamics: dQ = dU + dW = dU + PdV,


Where,


dQ = heat supplied,


dU = change in internal energy,


dW = PdV = work done, where P = pressure, dV = change in volume.


Since volume is constant, dV = 0 => dW = 0.


Hence, dQ = dU.


Heat(dQ) = 2 cal = nCvdT = 0.04 mol x Cv x 1K ,


Where n = number of moles, Cv = specific heat at constant volume, dT = rise in temperature.


=> Cv = 50 cal/mol/K. = (50 x 4.2 × 107)erg/cal x cal/mol/K = 2.1 x 109 erg/mol/K = 210 J/mol/K,


Since heat(J) = mechanical equivalent of heat x heat(cal) = 4.2 x heat(cal), and


1 J = 107 erg


We know, Cp = (Cv+R),


Where Cp = specific heat at constant pressure, Cv = specific heat at constant volume, R = universal gas constant = 8.314 J/kg/mol


Therefore, Cp = (210+8.314) J/mol/K = 218.314 J/mol/K.


Therefore, heat required to raise the temperature by 1°C at constant pressure = nCpdT,


Where n = number of moles, Cp = specific heat at constant pressure, dT = rise in temperature.


Hence, substituting, heat = (0.04 x 218.314 x 1) J = 8.737 J = (8.737/4.2) cal = 2.08 cal (since 1 J = 4.2 cal)(Ans)


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