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प्रश्न
Derive Mayer’s relation.
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उत्तर
- Consider one mole of an ideal gas that is enclosed in a cylinder by a light, frictionless airtight piston.
- Let P, V, and T be the pressure, volume, and temperature respectively of the gas.
- If the gas is heated so that its temperature rises by dT, but the volume remains constant, then the amount of heat supplied to the gas (dQ1) is used to increase the internal energy of the gas (dE). Since the volume of the gas is constant, no work is done in moving the piston.
∴ dQ1 = dE = CV dT ..............(1)
where CV is the molar specific heat of the gas at constant volume. - On the other hand, if the gas is heated to the same temperature, at constant pressure, the volume of the gas increases by an amount say dV. The amount of heat supplied to the gas is used to increase the internal energy of the gas as well as to move the piston backward to allow expansion of gas. The work is done to move the piston dW = PdV.
∴ dQ2 = dE + dW = Cp dT ..............(2)
Where CP is the molar specific heat of the gas at constant pressure. - From equations (1) and (2),
∴ Cp dT = CV dT + dW
∴ (Cp - Cv)dT = PdV ..............(3) - For one mole of gas,
PV = RT
∴ P dV = R dT, since pressure is constant.
Substituting equation (3), we get
(Cp - Cv) dT = R dT
∴ Cp - Cv = R
This is known as Mayer’s relation between CP and CV. - Also, CP = M0SP and CV = M0SV, where M0 is the molar mass of the gas and SP and SV are respective principal specific heats. Thus, M0SP - M0SV = R/J
Where J is the mechanical equivalent of heat.
SP - Sv = `"R"/("M"_0"J")`
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