Problem 1 : The gure below shows a cyclic thermodynamic process followed by n =

Question

Problem 1 : The gure below shows a cyclic thermodynamic process followed by n =

1:00 mol of a monatomic ideal gas. The initial pressure is p1 = 2:2  105 N=m2 and the

initial temperature is T1 = 560 K. The process 1 ! 2 is isothermal, the process 2 ! 3 is

constant pressure with nal temperature T3 = 360 K, and the process 3 ! 1 is adiabatic.

i) Determine the pressure p2 (in N=m2

), and the volumes V1; V2; V3 (in L). (ii) How much

work (W12,W23,W31) is done by the gas during each of the three segments? (iii) What is the

change in internal energy (E12,E23,E31) during each of the three segments?

The gure below shows a cyclic thermodynamic process followed by n = 1:00 mol of a monatomic ideal gas. The initial pressure is p1 = 2:2 105 N=m2 and the initial temperature is T1 = 560 K. The process 1 ! 2 is isothermal, the process 2 ! 3 is constant pressure with nal temperature T3 = 360 K, and the process 3 ! 1 is adiabatic. Determine the pressure p2 (in N=m2), and the volumes V1; V2; V3 (in L). (ii) How much work (W12,W23,W31) is done by the gas during each of the three segments? (iii) What is the change in internal energy (E12,E23,E31) during each of the three segments?

Explanation / Answer

Given monoatomic ideal gas

therefore Cp/Cv = 5/3 = 1.66 = gamma

P1*V1 = n*R*T

P1 = 2.2*(10^5) N/m^2

T1 = 560K = T2

n = 1 mol.

R = 8.314 J/K*mol

V1 = 8.314 * 1 * 560 / 2.2*(10^5)

V1 = 2.1163 * (10^-2) m^3 = 21.16 L

T3 = 360K

process 3-1 is an adiabiatic process

For adiabatic process

P * (V^gamma) = constant

(P^(1-gamma/gamma)) * T = constant

(P1^-0.4) * T1 = (P3^-0.4) * T3

T1/(P1^0.4) = T3/(P3^0.4)

P3 = 0.3313 * P1

P3 = 0.7288*(10^5) N/m^2

P1 * (V1^1.66) = P3 * (V3^1.66)

(V3^1.66) = P1 * (V1^1.66) / P3

V3 = 1.94 * V1

V3 = 4.1056 * (10^-2) m^3 = 41.05 L

P2 = P3 since 2-3 is a isobaric( constant pressure process )

P2 = 0.7288*(10^5) N/m^2

2-3 is a constant pressure process therefore

V / T = constant

V2 / T2 = V3 / T3

V2 = V3*T2/T3

V2 = 63.85 L = 6.385 * (10^-2) m^3

Part 2 :

For isothermal process 1-2

Work done by the gases(W1) = n * R * T * ln(V2/V1)

W12 = 1 * 8.314 * 560 * ln ( 63.85 L/21.16 L )

W12= 5142.02 J

For isobaric process 2-3

Work done by the gases(W2) = P2*(V3-V1)

W23 = 0.7288*(10^5) * (4.1056 * (10^-2) - 6.385 * (10^-2))

W23 = - 1662.8 J

here the negative sign indicates that work is done on the gas

For adiabatic process 3-1

Work done by the gases(W3) = [P1*V1 - P3*V3]/1-gamma = n * R * (T3 - T1)/1-gamma

W31 = 1 * 8.314 * (360-560) / 1-1.666

W31 = 2494.2 J

Part 3 :

In a cyclic process the net change in internal energy of the cycle is zero as the processes reach the initial state.

E12 + E23 + E31=0

For isothermal process 1-2

In an isothermal thermal process there is no change in temperature and since internal energy depends only on temperature therefore there is no change in internal energy in an isothermal process.

Change in internal energy E12 = 0 J

For adiabatic process 3-1

In an adiabatic process heat absorbed or released from the system is zero

Heat supplied to the gas = work done by the gas + change in internal energy

Q31 = W31 + E31

Q31 = 0

E31 = - W31

Change in internal energy E31 = -2494.2 J

Here the internal energy decreases of the gas decreases.

For isobaric process 2-3

E12 + E23 + E31=0

0 + E23 - 2494.2 = 0

Change in internal energy E23 = 2494.2 J

Please rate my answer and please comment if u have any doubt sir.

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