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.
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
1. Ideal gas equation: PV = nRT.
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,
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)