What wouldHow much work is done on the gas during each of the three segments? you like to ask?

If i get right you have the following changes

process 1: isochoric change from state 1 to state 2

process 2: isothermal change from state 2 to state 3

process 3: isobaric change from state 3 back to state1

Generally the work done on the gas is given by the integral:

W = - ∫ p dV from initial to final volume of the process

First process is isochoric, that means the volumes does not change. So the integral is zero and no work is done

W₁ = 0

Second process is isothermal. From Ideal gas law follows that

p∙V = n∙R∙T = constant

because n and T are constant.

Hence you may use either

p∙V = p₂∙V₂

or

p∙V = p₃∙V₃

to substitute p in the work integral.

W₂ = - ∫ p dV from V₂ to V₃

= - ∫ p₂∙V₂/V dV from V₂ to V₃

= - p₂∙V₂ ∙ ∫ 1/V dV from V₂ to V₃

= - p₂∙V₂ ∙ln( V₃/V₂ )

= - 4.16∙101325Pa ∙ 1×10⁻³m³ ∙ln( 5×10⁻³m³/1×10⁻³m³ )

= - 4.16∙101325Pa ∙ 1×10⁻³m³ ∙ln( 5×10⁻³m³/1×10⁻³m³ )

= -678.4J

(note: 1Pam³ = 1J)

Third process is isobaric, i.e

p = constant = p₃ = p₁

=>

W₃ = - ∫ p₁ dV from V₃ to V₁

= - p₁ ∙ ∫ dV from V₃ to V₁

= - p₁ ∙ (V₁ - V₃)

= p₁ ∙ (V₃ - V₁)

= 0.833∙101325Pa ∙ (5×10⁻³m³ - 1×10⁻³m³)

= 337.6J

To find the amount of heat transferred in each process, consider the change of internal energy. Change of internal energy equals heat transferred to the gas plus work done on it:

∆U = Q + W

Moreover the change of internal of an ideal gas is given by its change in temperature:

∆U = m∙Cv∙∆T

Hence:

Q = m∙Cv∙∆T - W

Mass specific heat capacity of Helium is

Cv = Cv,m / M = (3/2)∙R/M

= (3/2)∙8.314472J/molK / 4g/mol

= 3.118J/gK

Q₁ = m∙Cv∙(T₂ - T₁) - W₁

= 100×10⁻³g ∙ 3.118J/gK ∙ (1760 - 133)K - 0J

= 507.3J

Q₂ = m∙Cv∙(T₃ - T₂) - W₂

= 100×10⁻³g ∙ 3.118J/gK ∙ (1760 - 1760)K - (-678.4J)

= 678.4J

Q₃ = m∙Cv∙(T₁ - T₃) - W₃

= 100×10⁻³g ∙ 3.118J/gK ∙ (133 - 1760)K - 337.6J

= -844.9J

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