Thermodynamic Process

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a month ago

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alok kumar singh

Contributor-Level 10

Heat lost by steam =  Heat gained by water and calorimeter.

m * 540 + m * 1 * ( 100 - 31 ) = 200 540 m + 69 m = 1200 m = 1200 609 2

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a month ago

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

In isothermal process, temperature is constant.

In isochoric process, volume is constant.

In adiabatic process, there is no exchange of heat.

In isobaric process, pressure is constant

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

Na -> Avogadro's Number

J β 2 [ μ k R ] β 2 [ k R ] β [ k ] [ M L 2 T 2 K 1 ]

, and

-> S α 2 β [ M L 2 T 2 K 1 ] α 2 [ M L 2 T 2 K 1 ] α [ M 0 L 2 T 0 K 0 ] Dimensionless

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2 months ago

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A
alok kumar singh

Contributor-Level 10

Heat absorbed in cyclic process = Work done = 100? Joule

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alok kumar singh

Contributor-Level 10

Since process is isochoric, so

Q = n C v Δ T = n ( f 2 R ) Δ T = 4 ( 5 2 * R ) ( 5 0 ) = 5 0 0 R

New answer posted

2 months ago

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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

2 months ago

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alok kumar singh

Contributor-Level 10

Kindly go through the solution 

  T 1 4 0 0 = 3 0 0 2 4 0 T 1 = 5 0 0 K

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