Ideal gas Cp, Cv, equation of state, 1^st law of Thermodynamics. mole of classic

Question

Ideal gas Cp, Cv, equation of state, 1^st law of Thermodynamics. mole of classical noninteracting (ideal) diatomic gas is placed in a tank sealed with a moving, friction-free, leak-free piston. The piston faces outside air, so the pressure in the tank is kept at P_o = 1 atm. The initial temperature T_o=300 K. calculate the volume of the gas in the initial state, V_o, in liters, lm^3 =1000 1. The gas is then cooled until its volume becomes 1/2 V_o. How much energy, in Joules, had to be removed from the gas?

Explanation / Answer

Applying Gas equation,

PV = n R T

(1 x 101325 Pa) ( V ) = (1 mole) (8.314) ( 300 K )

V = 0.0246 m^2

V( in litre) = 24.62 L ............Ans(a)

b) Pressure and n will be constant.

hence at work done work done on gas , W = P deltaV = 101325 ( 0.0246 - 0.0246/2)

W = 1246.3 J .......Work done gas by outside.

Or energy removed will be = n Cp deltaT

for diatomic gas, Cp = 7R/2

and V / T = constant

V is halved so T also halved.

Tf = 150 K

Q = 1 x 7 x 8.314/2 x (300 - 150) = 4364.85 J ......energy removed.

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