Thermodynamics

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A
alok kumar singh

Contributor-Level 10

Efficiency η = Q_net/Q? = (1915-40+125-Q)/ (1915+125) = 0.5.
(2000-Q)/2040=0.5 ⇒ Q=980J.

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A
alok kumar singh

Contributor-Level 10

(f? n? RT? )/2 + (f? n? RT? )/2 = 3/2 RT + 5/2 * 3RT = 15

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V
Vishal Baghel

Contributor-Level 10

T? V? ¹ = T? V? ¹
T? = T? (V? /V? )? ¹ = T? (10)? /? ¹ = T? (10)²/?
ΔV = 5/2 nR; 5/2 * 5 * 3102/5-1
625/6 * 1.5 * 293 = 461440 ≈ 46ks

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Vishal Baghel

Contributor-Level 10

n' = 0.75n? +0.5n? = 1.25n? moles. P? /P? =5.

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V
Vishal Baghel

Contributor-Level 10

f=5
∴ U = (5/2)RT
And γ = 1+2/f = 1+2/5=7/5

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V
Vishal Baghel

Contributor-Level 10

13.49
2 A (g) → A? (g)
Δn_g = 1 - 2
Δn_g = 1
ΔH = ΔU + Δn_gRT
ΔH = -20 * 10³ + (-1)8.31 * 298
ΔH = -22477.572 J mol? ¹
ΔG = ΔH - TΔS
ΔG = -22477.572 - (-30)298 = -22477.572 + 8940
ΔG = -13537.57 J mol? ¹

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a month ago

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V
Vishal Baghel

Contributor-Level 10

In adiabatic process
PV? = constant
P (m/ρ)? = constant
As mass is constant
P ∝ ρ?
P_f/P_i = (ρ_f/ρ_i)? = (32)? /? = 2? = 128

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Vishal Baghel

Contributor-Level 10

  Δ l = 0.03 m m

P 1 = 1 a t m , T 1 = 273 K

Now work done  P 1 V 1 γ = P 2 V 2 γ P 2 = P 1 V 1 V 2 γ = 1 a t m 1 3 1.4

Closes answer is  = P 1 V 1 - P 2 V 2 γ - 1 = 88.7 J .

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Vishal Baghel

Contributor-Level 10

  γ mixture   = n 1 C P 1 + n 2 C P 2 n 1 C V 1 + n 2 C V 2 = n 1 γ 1 R γ 1 - 1 + n 2 γ 2 R γ 2 - 1 n 1 R γ 1 - 1 + n 2 R γ 2 - 1

on rearranging we get

n 1 + n 2 γ mix   - 1 = n 1 γ 1 - 1 + n 2 γ 2 - 1 ; 5 γ mix   - 1 = 3 1 / 3 + 2 2 / 3

5 γ mix   - 1 = 9 + 3 = 12 γ mixture   = 17 12 + 1 + 5 12 ; γ mix   = 1.42

 

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