Thermodynamics

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New answer posted

3 weeks ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

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V
Vishal Baghel

Contributor-Level 10

Δ U = 7 4 2 . 2 4 k J / m o l e Δ H 2 9 8 = ?

NH2CN (s) + 3 2 O2 (g) ® N2 (g) + O2 (g) + H2O (l)

= 7 4 2 . 2 4 + 1 2 * 8 . 3 1 4 1 0 0 0 * 2 9 8

7 4 2 . 2 4 + 1 . 2 3 8 = 7 4 1 . 0 0 2 7 4 1 k J / m o l

So, magnitude of Δ H = 741 kJ/mol.

New answer posted

3 weeks ago

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V
Vishal Baghel

Contributor-Level 10

At constant pressure,

d U = n C V d T = n . 5 2 R d T

d Q = n C P d T = n . 7 2 R d T

dW = nRdT

dU : dQ : dW = 5 : 7 : 2

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Vishal Baghel

Contributor-Level 10

Moles of N2O= 2 . 2 4 4 = 1 2 0

Δ H = n C p Δ T = 1 2 0 * 1 0 0 ( 4 0 ) = 2 0 0 J

Δ U = q p + w

w = P e x t . Δ V

w = 1 ( 1 6 7 . 7 5 2 1 7 . 1 ) 1 0 0 0 * 1 0 1 . 3 J = + 5 J

Δ U = 2 0 0 + 5 = 1 9 5 J

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Vishal Baghel

Contributor-Level 10

Process-AB Isobaric,                                                           

Process-AC Isothermal, and

Process-ADÞAdiabatic

W2 < W1 < W3

 

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A
alok kumar singh

Contributor-Level 10

Process-AB Isobaric,                                                           

Process-AC  Isothermal, and

Process-ADÞAdiabatic

W2 < W1 < W3

 

New answer posted

3 weeks ago

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A
alok kumar singh

Contributor-Level 10

U = a P 4

P V γ - 1 = a P 4

P - 3 V = a γ - 1

P V - 1 / 3 = c o n s t a n t

C o m p a r i n g w i t h P V x = c o n s t a n t

x = - 1 3

C = R γ - 1 + R 1 - x = R 5 3 - 1 + R 1 + 1 3 = 3 R 2 + 3 R 4

C = 9 R 4

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Vishal Baghel

Contributor-Level 10

q = 0, w = 0, U = 0 = f (V, T)

As volume changes, temperature must change to maintain constant internal energy.

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Vishal Baghel

Contributor-Level 10

(0) Path A to D is adiabatic.                                                                                                                                              

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V
Vishal Baghel

Contributor-Level 10

msx (16 – 10) = msy (20 – 16)

s x s y = 2 3

m s y ( 2 6 2 0 ) = m s z ( 3 0 2 6 )

s y s z = 2 3

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