A) - 300 joule
B) +100 joule
C) - 100 joule
D) None of these
Correct Answer: B
Solution :
Given that, \[q=200\,J\] \[\Delta V=500\,c{{m}^{3}}=500\times {{10}^{-6}}{{m}^{3}}\] \[P=2\times {{10}^{5}}N{{m}^{-2}}\] According to 1st law of thermodynamics \[\Delta E=q-P\Delta V\] \[=200-(500\times {{10}^{-6}}\times 2\times {{10}^{5}})\] \[=200-100=100\,\text{joule}\]You need to login to perform this action.
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