Question
The average translational energy and the $$rms$$ speed of molecules in a sample of oxygen gas at $$300\,K$$ are $$6.21 \times {10^{ - 21}}J$$ and $$484\,m/s$$ respectively. The corresponding values at $$600\,K$$ are nearly (assuming ideal gas behavior)
A.
$$12.42 \times {10^{ - 21}}J,968\,m/s$$
B.
$$8.78 \times {10^{ - 21}}J,684\,m/s$$
C.
$$6.21 \times {10^{ - 21}}J,968\,m/s$$
D.
$$12.42 \times {10^{ - 21}}J,684\,m/s$$
Answer :
$$12.42 \times {10^{ - 21}}J,684\,m/s$$
Solution :
The formula for average kinetic energy is
$$\eqalign{
& \overline {K.E} = \frac{3}{2}kT \cr
& \therefore \frac{{{{\left( {\overline {K.E} } \right)}_{600\;K}}}}{{{{\left( {\overline {K.E} } \right)}_{300\;K}}}} = \frac{{600}}{{300}} \cr
& \Rightarrow {\left( {\overline {K.E} } \right)_{600\;K}} = 2 \times 6.21 \times {10^{ - 21}}\,J \cr
& = 12.42 \times {10^{ - 21}}\,J \cr} $$
Also the formula for $$r.m.s.$$ velocity is
$$\eqalign{
& {C_{rms}} = \sqrt {\frac{{3KT}}{m}} \cr
& \therefore \frac{{{{\left( {{C_{ms}}} \right)}_{600\;K}}}}{{{{\left( {{C_{rms}}} \right)}_{300\;K}}}} = \sqrt {\frac{{600}}{{300}}} \cr
& \Rightarrow {\left( {{C_{rms}}} \right)_{600\;K}} = \sqrt 2 \times 484 = 684\,m/s \cr} $$