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Two moles of helium are initially at a temperature of 27.0 ∘C and occupy a volume...

Two moles of helium are initially at a temperature of 27.0 ∘C and occupy a volume of 3.20×10−2 m3 . The helium first expands at constant pressure until its volume has doubled. Then it expands adiabatically until the temperature returns to its initial value. Assume that the helium can be treated as an ideal gas.

A)What is the total heat supplied to the helium in the process? In Joules

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.

B)What is the total change in internal energy of the helium? In Joules

.

.

C)What is the total work done by the helium? In Joules

.

.

D)What is the final volume of the helium? In m3

.

.

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Answer #1

The system undergoes two processes, Isobaric Expansion and Adiabatic expansion. Since in Adiabatic process, there no heat transfer between the system and surrounding, the total heat supplied to the system is equal to the heat supplied in Isobaric expansion. Heat supplied to the system in isobaric process is, The initial pressure can be calculated by nRT 2x8.314x300 3.2x10 P-155887.5Pa Change in Volume 2(3.20x102)-(3.20x102) 3.20x 10-2㎡ The heat supplied to the system is ADIAB ATIC N m AQ-x2molx8.314x(600K-300K) molK AQ-12.5x10J

b) Change in Internal energy = Δυ lSOBA RIC Change in internal energy in Isobaric Process: AU-AQ-WD Work Done is - WD-PAV + Δυ.Dunc ADIABATIC -155887.5Pax3.20x10 m Therefore, change in internal energy in Isobaric Process AU 12.5x10 J-5x10 J Δυ 7.5x10;J Change of Internal energy in Adiabatic Process Work done in adiabatic process is given by (y-1) WD=2x3.3 14x(300) 166-1 WD-7.5x10J

Therefore Ar AU-7.5x10 Net Change in Internal Energy 彫:Nisanser NU.asa u c c) Total Work done =5x10,I+7.5x10 =12.5×101

d) The final process is adiabatic process, For adiabatic process, the equation can be V T,11.66-1 V, (T, T,-600 K V,-6.40 x 10m3 T,-300 K 1.66-1 V,-0.064 (2)RĪ V,-0.1829m,

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