##### 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)

1