Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are labeled as A → B, A → C and A → D. The change in internal energies during these process are taken as EAB, EAC and EAD and the work done by WAB, WAC and WAD. The correct relation between these parameters are:

 

Option 1 -

EAB < EAC < EAD, WAB > 0, WAC > WAD

Option 2 -

EAB = EAC = EAD, WAB > 0, WAC = 0, WAD > 0

Option 3 -

EAB > EAC > EAD, WAB < WAC < WAD

Option 4 -

EAB = EAC < EAD, WAB > 0, WAC = 0, WAD < 0

0 2 Views | Posted a month ago
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A
alok kumar singh

Kindly go through the solution

 

A
alok kumar singh

From first law of thermodynamics

? W = ? Q - ? U

? Q = n C p ? T

? U = n C v ? T

? W = n ( C p - C v ) ? T

? Q ? W = γ γ - 1 C p C p = γ

A
alok kumar singh

From first law of thermodynamics

? W = ? Q - ? U

? Q = n C p ? T

? U = n C v ? T

? W = n ( C p - C v ) ? T

? Q ? W = γ γ - 1 C p C p = γ

A
alok kumar singh

Δ Q = Δ U + Δ W

Q = Δ U + Q 5 Δ U = 4 Q 5 = n C v Δ T 4 Q 5 = 5 R 2 Δ T Δ T = 8 Q 2 5 R

Q = n c Δ T = 1 × C × 8 0 2 5 R C = 2 5 R 8 x = 2 5

A
alok kumar singh

Given                                                                           

QAB = +40J

QBC = 0J

QCA = -60J

WCA =?

WBC = -50J (on the gas)

UA = 1560J

1st Law on path A to B

WAB +   Δ U = Q

Þ 0 +    [ U B U A ] = 4 0

Þ UB = UA + 40

= 1560 + 40

UB = 1600 J

 1st Low

...more

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