P20 086

background image

86.

(a) The temperature is 10

C

→ T = 283 K. Then, with n = 3.5 mol and V

f

/V

0

= 3/4, we use Eq. 20-14:

W = nRT ln



V

f

V

0



=

2369 J ≈ −2.4 kJ .

(b) The internal energy change ∆E

int

vanishes (for an ideal gas) when ∆T = 0 so that the First Law

of Thermodynamics leads to Q = W =

2.4 kJ. The negative value implies that the heat transfer

is from the sample to its environment.


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