A) 1200 J
B) 1300 J
C) 1400 J
D) 1500 J
Correct Answer: C
Solution :
Given that: \[{{C}_{p}}=7\,J{{K}^{-1}}mo{{l}^{-1}}\] Weight of\[{{N}_{2}}\]gas = 28 gm \[{{T}_{1}}=0{}^\circ C=273+0=273\text{ }K\] \[{{T}_{2}}=100{}^\circ C=273+100=373\text{ }K\] Now, by using: Number of moles \[=\frac{Weight\text{ }of\text{ }{{N}_{2}}}{Molecular\text{ }weight\text{ }of\text{ }{{N}_{2}}}=\frac{28}{14}=2\] Now, by using \[\Delta H=n{{C}_{p}}({{T}_{2}}-{{T}_{1}})=2\times 7(373-273)\] \[=14\times 100=1400\text{ }Joule\]
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