an ideal gas initally at 600 K and 10 bar undergoes a four step mechanically rev

Question

an ideal gas initally at 600 K and 10 bar undergoes a four step mechanically reversible cycle in a closed system. In step 1-2, pressure decreases isothermally to 3 bar; in step 2-3, pressure decreases at a constant volume to 2 bar; in step 3-4, volume decreases at constant pressure, and in step 4-1 the gas returns adiabatically to its initial state. Take Cp = (7/2)R and Cv = (5/2)R. a) determine (where unknown) both T and P for states 1,2,3,4 ..... b) calculate Q W dU and dH for each step of the cycle

Explanation / Answer

step 1-2:

T1 = 600 K and P1 = 10 bar

P2 = 3 bar

and T2 T1 = 600 K (isothermal)

dU = 0 (as T = consatnt)

dW = R.Tln(P1/P2) = R x 600 x ln (10/3) = 721.78R

Q = dW = 721.78 R

dH = 0 (T = constant)

stwp 2-3 :

P3 = 2 bar

P2/T2= P3/T3

so, 3/600 = 2/T3

so, T3 = 400 K

W = 0

Q = dU = Cv.dT = (5/2)R x (400 - 600) = - 500 R

dH = Cp.dT = (7/2)R (400 - 600) = - 700 R

step 3-4 :

P4 = P3 = 2 bar

also T4/T1 = (P4/P1)^(y-1)/y

y = 1.4 (Cp/Cv)

so, T4/600 = (2 /10)^(1.4 -1)1.4

so,T4 = 492.2 K

so for 3-4

dU = Cv.dT = (5/2)R x (492.2 - 400) = 230.5 R

Q = Cp.dT = (7/2)R x (492.2 - 400) = 322.7 R

W = Q - dU = 322.7R - 230.5R = 92.2 R

H = Cp.dT = (7/2)R x (492.2 - 400) = 322.7 R

for 4-1

Q = 0 (as adaibatic)

W = R (T4 - T1)/(y-1) = R(492.2 - 600)/(1.4-1) = - 269.5 R

dU = - W = 26.5R

H = Cp.dT = (7/2)R x (492.2 - 600) = - 377.5R

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