Derive Meyer's relation.

1.	Derive Meyer's relation.
Answer» Solution :Consider are mole of an ideal gas enclosed in a cylinder provided with a frictionless piston of area A . P-pressure of gas V- volume of gas T - absolute temperature gas dQ- quantity of heat supplied To keep the volume of the gas constant a small `w_t` is placed over the piston. The pressure and temperature increase to p + dp and T + dt. dQ is used to increase the internal ENERGY du of the gas. But the gas does not do any work [dw= 0] `therefore dQ = dU = 1 xx c_v xx dT`. Now the w, is removed. The piston now moves upwards thus'a dist. DX, the pres. of the enclosed gas equal to atmosphere pressure P. Due to expansion, temperature decreases. Now a quantity of heat dQ’ is supplied till its temperature become `T+DeltaT.` This heat energy is not only used to increase the internal energy dU of the gas but also to do exists work dW in moving the piston upwards. ` therefore dQ^1 =dU + dW` At constant pressure ` dQ^1 = c_p = dT` ` therefore c_p dT = c_v = c_vdT + dW` work done dW = Force x dist. `P xx A xx dx` ` dW = P.dv` [A. dx = dv change in volume] ` therefore c_p dT = c_v dT + Pdv`...(1) The EQN of state of an ideal gas is ` PV = RT` Difference both the sides pdv = Rdt...(2) Subtract (2) in (1) `c_p dT = c_v dT + RdT ,` `c_p = c_v + R ""therefore c_p - c_v + R` This equation is known as Meyer's electron.

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