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Physics Question on kinetic theory

Heat absorbed by a mass of 1 g of helium, when its temperature rises from 11∘C11{}^\circ C to 131∘C131{}^\circ C at constant volume, is H. Heat absorbed by 7 g of nitrogen, when its temperature rises from 11∘C11{}^\circ C to 71∘C71{}^\circ C at constant volume, is HN{{H}_{N}} . The ratio of H to HN{{H}_{N}} is

A

32\frac{3}{2}

B

43\frac{4}{3}

C

65\frac{6}{5}

D

87\frac{8}{7}

Answer

65\frac{6}{5}

Explanation

Solution

Heat absorbed at constant volume H=nCVΔTH=n{{C}_{V}}\Delta T where, n=n= number of moles =mM=\frac{m}{M} Heat absorbed by helium H=14×32R(131−11)H=\frac{1}{4}\times \frac{3}{2}R(131-11) =14×32R×120=\frac{1}{4}\times \frac{3}{2}R\times 120 =45R=45R [For the (monoatomic gas), CV=32R{{C}_{V}}=\frac{3}{2}R ] Heat absorbed by nitrogen HN=728×52R(71−11){{H}_{N}}=\frac{7}{28}\times \frac{5}{2}R(71-11) =14×52R×60=\frac{1}{4}\times \frac{5}{2}R\times 60 =752R=\frac{75}{2}R [For H2{{H}_{2}} (diatomic gas) CV=52R{{C}_{V}}=\frac{5}{2}R ] HHN=45R(75/2)R=65\frac{H}{{{H}_{N}}}=\frac{45R}{(75/2)R}=\frac{6}{5}