Question

During process A → B is shown in figure. The change
The P-V curve for 1 mole of an ideal monoatomic gas during process A - B is shown in
in internal energy of the gas is-
(Pressure goes from P to 2P and Volume goes from V to 2V in a parabolic path)​

Answer 1

Answer:

The given process follows a linear P-V relation given by:

v−V

p−P

=

V−2V

P−P/2

=−P/2V

⇒p−P=

2V

−P

(v−V)⇒p=−

2V

Pv

+3P/2

Here work done ΔW=area under P-V diagram =3PV/4=0.75PV

For isotherm, pv=PV ⇒ work done ΔW=PVln2=0.693PV

Hence statement a is correct

For t-v diagram, replace p in (i) by

v

nRt

Hence, t=−

2nRV

Pv

2nR

3Pv

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Answer 2

Given info : The P-V curve for 1 mole of an ideal monoatomic gas during process A → B is shown in figure.

To find : The change in internal energy of the gas is ...

solution : we have to find the temperatures at point A and B, after that we can find change in internal energy.

at point A,

from figure , P₀V₀ = nRTA

⇒P₀V₀ = 1 × R × TA

⇒TA = P₀V₀/R ......(1)

at point B,

from figure, 2P₀ × 2V₀ = 1 × R ×TB

⇒TB = 4P₀V₀/R .......(2)

from equations (1) and (2) we get,

change in temperature, ∆T = TB - TA = 3P₀V₀/R

now change in internal energy , U = nCv∆T

= 1 × 3R/2 × 3P₀V₀/R [ for monoatomic, Cv = 3R/2 ]

= 9P₀V₀/2

Therefore the change in internal energy is 9P₀V₀/2