First slide

Change in internal energy

Question

6 moles of oxygen, 4 moles of Ar. and 2 moles of water vapour form an ideal gas mixture. The mixture is made to expand adiabatically from an initial temperature 127°C to a final temperature 27°C, decrease of internal energy of the mixture is: (take R=2 cal/mol-K)

Moderate

A

3600 calorie
B

4800 calorie
C

5400 calorie
D

6600 calorie

Solution

Decrease in internal energy $= -∆ U$
$= (n_{1} C_{V_{1}} + n_{2} C_{V_{2}} + n_{3} C_{V_{3}}) (T_{1} - T_{2}) = (6 \times \frac{5}{2} R + 4 \times \frac{3}{2} R + 2 \times 3 R) (100) = 2700 R = 5400 cal$

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