Question

A 10 L adiabatic container contains 0.1 mol of air, at a temperature of 250 K. It is closed by a stopcock. Ambient air is at 300 K at a pressure of 1 bar. The stopcock is opened for a moment and a certain amount of air penetrates into the container. After thermal equilibrium is reached inside the container, the gas temperature is measured to be 350 K. 1) Assuming that the air behaves as an ideal gas, find the initial pressure in the container, the number of moles of air that penetrated in the container, the sum of the internal energy changes for the gas inside the container and the gas that entered it and the final pressure inside. 2) Find the enthalpy and entropy changes of the gas that is inside the container at the end of the process. Use κ = Cp,mol/CV,mol = 7/5 and Cp,mol − CV,mol = R.

Answer 1

P - pressure, V- Volume, T - temperature,

1, 2 , 0 are initial , final and ambient condition respectively.

let P1 is the initial pressure of an adiabatic container

Other Initial properties : V1 = 10 L = 0.01 m3; n1 = 0.1 mol; T1 = 250 K

according to Ideal gas equation P1 *V1 = n1*R*T1

P1 * 0.01 = 0.1 * 8.314 * 250

initial pressure = P1 = 20785 N/m²

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Here total energy inside the container is always internal energy.

While the energy assosiated with enetering air is expressed as enthalpy.

m- mass, u- specific internal energy, h - specific enthalpy

according to mass conservation: m2 - m1 = mi.................................................................(1)

according to energy conservation m2u2 - m1u1 = mih0.......................................................(2)

from equation (1) and (2)

m2u2 - m1u1 = (m2-m1)h0

dividing whole equation by atomic mass of air (m/M = n or mass / atomic weight = no. of mole)

n2u2 - n1u1 = (n2-n1)h0

n2* Cv * T2 - n1 * Cv * T1 = (n2 - n1) * Cp * T0

n2* T2 - n1 * T1 = (n2 - n1) *( Cp /Cv)* T0

n2*350 - 0.1*250 = (n2-0.1)*(7/5)*300

350 * n2 - 25 = 420 *n2 - 42

70* n2 = 17

n2 = 0.243

number of moles of air that penetrated in the container = final no. of mole - initial no. of mole

= n2 - n1 = 0.243 - 0.1 = 0.143

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Cp / Cv = 7/5 ; Cp = 7/5 Cv

Cp - Cv = R = 8.314 J/mol/K

7/5 Cv - Cv = R

Cv (7/5 -1) = R

Cv * 2/5 = R

Cv = 5R/2 = 5 * 8.314 / 2 = 20.785 J/mol/K

Cp = 7/5 * 20.785 = 29.1 J/mol/K

change in internal energyof gas inside (initial) the container = n1 * Cv * (T2 -T1)

= 0.1 * 20.785 * (350 - 250 ) = 207.85 J

change in internal energy of gas entered in the container = (n2-n1) * Cv * (T2 -T0)

= 0.143 * 20.785 * (350 - 300 ) = 148.61 J

sum of the internal energy changes for the gas inside the container and the gas that entered it

= 207.85 + 148.61 = 356.46 J

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According to ideal gas equation : P2 = n2* R * T2 / V2

P2 = 0.243 * 8.314 * 350 / 0.01 = 70710.57 N/m²

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change in enthalpy = change in internal energy + [n2* T2 - n1* T1 - (n2-n1)*T0]*R

= 356.46 + [ 0.243 *350 - 0.1*250 - 0.143*300] * 8.314  

= 499 J

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Entropy change of inside air = n1 *(Cp ln T2/T1 - R ln (P2/P1))

= 0.1 * (29.1 * ln (350/250) - 8.314 * ln (70710.57/ 20785)) = -0.0388 J/K

Entropy change of entered air = (n2-n1) *(Cp ln T2/T0 - R ln (P2/P0))  

= 0.143 * (29.1 * ln (350/300) - 8.314 * ln (70710.57 / 10⁵)) = 1.052 J/K

Total change in entropy of air inside tank = -0.0388 + 1.052 = 1.0132 J/K