If i get right you have the following changesprocess 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 state1Generally the work done on the gas is given by the integral:W = - ∫ p dV from initial to final volume of the processFirst process is isochoric, that means the volumes does not change. So the integral is zero and no work is doneW₁ = 0Second process is isothermal. From Ideal gas law follows thatp∙V = n∙R∙T = constant because n and T are constant.Hence you may use either p∙V = p₂∙V₂ orp∙V = p₃∙V₃to 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.ep = 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)= 337.6JTo 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 + WMoreover the change of internal of an ideal gas is given by its change in temperature:∆U = m∙Cv∙∆THence:Q = m∙Cv∙∆T - WMass specific heat capacity of Helium isCv = Cv,m / M = (3/2)∙R/M = (3/2)∙8.314472J/molK / 4g/mol= 3.118J/gKQ₁ = m∙Cv∙(T₂ - T₁) - W₁= 100×10⁻³g ∙ 3.118J/gK ∙ (1760 - 133)K - 0J= 507.3JQ₂ = m∙Cv∙(T₃ - T₂) - W₂= 100×10⁻³g ∙ 3.118J/gK ∙ (1760 - 1760)K - (-678.4J)= 678.4JQ₃ = m∙Cv∙(T₁ - T₃) - W₃= 100×10⁻³g ∙ 3.118J/gK ∙ (133 - 1760)K - 337.6J= -844.9J...
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