A) 0.25
B) 0.75
C) 1
D) 0.67
Correct Answer: B
Solution :
Initial internal energy, \[{{U}_{1}}=n{{C}_{\upsilon }}{{T}_{1}}\] Final internal energy, \[{{U}_{2}}=n\,{{C}_{v}}{{T}_{2}}\] \[\therefore \]Fractional change \[=\frac{{{U}_{1}}-{{U}_{2}}}{{{U}_{1}}}=\frac{n{{C}_{\upsilon }}({{T}_{1}}-{{T}_{2}})}{_{n}{{C}_{\upsilon }}{{T}_{1}}}\] \[=1-\frac{{{T}_{2}}}{{{T}_{1}}}=1-\frac{273+27}{273+927}\] \[=1-\frac{300}{1200}\] \[=1-\frac{1}{4}=\frac{3}{4}=0.75\]You need to login to perform this action.
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