A) \[-2.16J\]
B) \[12.156\text{ }J\]
C) \[2.16J\]
D) \[101.3\text{ }J\]
Correct Answer: C
Solution :
Here: change in volume \[(V)=500-300=200\text{ }cc=0.2\text{ }L\] Pressure (P)=0.6 atm and heat liberated \[(q)=10\,J\] Work done \[(W)=P\Delta V=(0.2\times 0.6)=0.12L-\text{ }atm\] But \[1\text{ }L-atm=101.3\text{ }J\] Hence, work\[=0.12\times 101.3=12.156\text{ }J\]. We also know that heat is liberated. Therefore, it would be negative. Thus, change in energy \[\Delta E=q+W=-10+12.16=2.16\,J\]You need to login to perform this action.
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