Question

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Answer 1

Answer:

Explanation:

GivenCp= 5R/2, mole–1deg–1Cv= 3R/2, mole–1deg–1R=8.314J mole–1deg–1Step 1.To calculate the value of T2, the final temperature, using the equation(T2/T1)γ= (P2/ P1)γ– 153pvCCγ ==Substituting the value of γin (1)(T2 / 273.2)5/3=(2 / 20)Solving it, we getT2=108.8 KStep 2. To calculate maximum work under adiabatic conditions21max(–)1 –nR TTw=γ28.314 (108.8 – 273.2)1 – 5/3×==4100 J =4.1 kJ