A) - 2.16 J
B) 12.156 J
C) 2.16 J
D) 101.3 J
Correct Answer: C
Solution :
[c] Here: Change in |
Volume \[(V)=500-300=200cc=0.2\,litre,\] |
Pressure \[(P)=0.6\,atm\]and heat liberated (q) = 10J |
Work done \[(W)=P\Delta V=(0.2\times 0.6)=0.12\,litre-atm\] |
But \[1\,litre-atm=101.3\,J\]. |
hence \[W=0.12\times 101.3=12.156\,J\]. We also know that heat is liberated, therefore it would be negative. Thus change in \[\Delta E=q+W=-10+12.16=2.16\,J\]. |
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