Thermodynamics

Get insights from 323 questions on Thermodynamics, answered by students, alumni, and experts. You may also ask and answer any question you like about Thermodynamics

Follow Ask Question
323

Questions

0

Discussions

4

Active Users

0

Followers

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

V = k T 2 3

T V 3 2 = K

γ 1 = 3 / 2

γ = 1 / 2

Work done = n R Δ T y + 1 = 1 * R * 9 0 3 / 2 = 6 R

n = 60

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

U = 3PV + 4

n C v T = 3 P V + 4

n * f R T 2 = 3 P V + 4

f = 6 + 8 P V

f > 6 p o l y a t o m i c

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

n = 1 mole

W A B = n R T l n v 2 v 1 = 1 * R T l n 2 v 1 v 1 = R T l n 2

W B C = 0

W C A = P 1 V 1 P 2 V 2 γ 1 = P 1 4 * 2 V 1 P 1 * V 1 γ

W C A = R T 2 ( γ 1 )

W t o t a l = R T [ l n 2 1 2 ( γ 1 ) ]

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

P = kV3 Þ PV-3 = k Þ Polytropic Process with index m = -3

W = n R ( T 2 T 1 ) 1 m = n R ( 3 0 0 1 0 0 ) 1 ( 3 ) = 5 0 n R .

New answer posted

a month ago

0 Follower 2 Views

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

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Since process is isochoric

So    Δ U = n C v Δ T


Δ U = n ( 5 2 R ) Δ T ( i ) [ C V = 5 2 R ]      

And external work


Δ W = n R Δ T ( i i )    

5 2 = x 1 0 x = 2 5 . 0 0                

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

from newtan's law of cooling :

d T d t = k ( T T 0 )

Using average value :

7 5 6 5 5 = k [ ( 7 5 + 6 5 ) 2 2 2 5 ]

( 6 5 T ) 5 = k [ ( 6 5 + T ) 2 2 5 ]

1 0 6 5 T = 4 5 6 5 + T 2 2 5

T = 5 7 ° C

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Total energy of monoatomic gas at equilibrium

= 3 2 K B T

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

From   Δ H = Δ G + T Δ S Δ S = Δ H Δ G T

Δ S = [ 5 1 . 4 ( 4 9 . 4 ) 3 0 0 ] * 1 0 0 0 J K 1 m o l 1 = 3 3 6 J K 1 m o l 1

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Since process is isothermal, so

W = n R T l n ( V f V i ) = 1 * 8 . 3 * 3 0 0 * l n ( 4 2 ) = 1 * 8 . 3 * 3 0 0 * l n ( 2 )       

W = 1 * 8 . 3 * 3 0 0 * 0 . 6 9 3 1 1 7 2 5 . 8 2 J 1 7 2 5 8 * 1 0 1 J

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 687k Reviews
  • 1800k Answers

Share Your College Life Experience

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.