Question

Heat absorbed by a mass of 1 g of helium, when its temperature rises from $ 11{}^\circ C $ to $ 131{}^\circ C $ at constant volume, is H. Heat absorbed by 7 g of nitrogen, when its temperature rises from $ 11{}^\circ C $ to $ 71{}^\circ C $ at constant volume, is $ {{H}_{N}} $ . The ratio of H to $ {{H}_{N}} $ is

• A. $ \frac{3}{2} $
• B. $ \frac{4}{3} $
• C. $ \frac{6}{5} $
• D. $ \frac{8}{7} $

Answer 1

The Correct Option is C

Heat absorbed at constant volume $ H=n{{C}_{V}}\Delta T $ where, $ n= $ number of moles $=\frac{m}{M} $ Heat absorbed by helium $ H=\frac{1}{4}\times \frac{3}{2}R(131-11) $ $=\frac{1}{4}\times \frac{3}{2}R\times 120 $ $=45R $ [For the (monoatomic gas), $ {{C}_{V}}=\frac{3}{2}R $ ] Heat absorbed by nitrogen $ {{H}_{N}}=\frac{7}{28}\times \frac{5}{2}R(71-11) $ $=\frac{1}{4}\times \frac{5}{2}R\times 60 $ $=\frac{75}{2}R $ [For $ {{H}_{2}} $ (diatomic gas) $ {{C}_{V}}=\frac{5}{2}R $ ] $ \frac{H}{{{H}_{N}}}=\frac{45R}{(75/2)R}=\frac{6}{5} $