Question

Find the work done from the following graph. 92 = 600 Joule 500 K P(N/m2) 350 K 9, V>​

Answer 1

Answer:

n=3 moles, T=27

o

C=300K

P

1

=10atm P

2

=1atm

Work done in isothermal reversible process,

W=−2.303nRTlog(P

1

/P

2

)

=−2.303×3×8.314×300log(

1

10

)

=5744.14J

Work done against constant pressure of 1atm

=P

opp

(V

f

−V

i

)

=−1atm(

P

2

nRT

−

P

1

nRT

)

=−1×3×8.314×300(

1

1

−

10

1

)

=6734.34J .