A) 30
B) 50
C) 70
D) 90
Correct Answer: B
Solution :
Invoking first law of Thermodynamics for constant pressure process, we get \[\Delta Q=\Delta U+\Delta W\] \[70=\Delta U+nR\Delta T\] \[70=\Delta U+2\times 2\times 5\] \[\therefore \] \[\Delta U=70-20=50\,\]calories For constant volume process:\[\Delta Q=\Delta U+\Delta W\] \[\Delta U\]will remain 50 calories because the temperature change is same and \[\Delta W\]will be 0 for constant volume process. \[\therefore \]\[\Delta Q=50\]calories Hence, the correction option is [b].You need to login to perform this action.
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