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