A) \[3.775cal\]
B) \[37.256cal\]
C) \[372.56cal\]
D) \[3725.6cal\]
Correct Answer: C
Solution :
Key Idea: We know that \[{{C}_{P}}-{{C}_{V}}=R\] or \[{{C}_{P}}={{C}_{V}}+R\] \[\therefore \] \[{{C}_{V}}=\frac{3}{2}R\] \[\because \] \[{{C}_{P}}=\frac{3}{2}R+R=\frac{5}{2}R\] At constant pressure heat given to one mole gas \[\Delta H=m\,s\Delta T\] \[\therefore \] \[\Delta H={{q}_{p}}=1\times \frac{5}{2}R\times (373-298)\] \[=1\times \frac{5}{2}\times 1.987\times 75\] \[=372.56cal\]
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