One mole of oxygen occupies a volume of 24 L at 20- C and 1 atm presssure- If the total momentum of the molecules in kgm-s is 10-x-then the value of x is

Root mean square speed of the gas= √(3RTM_0) =√(3× R× 29332× 10^ 3)=477.89m/s Thus the total momentum of the molecules= mv_rms=32× 10^ 3× 477.89kgm ...

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Standard XII

Physics

Temperature

One mole of o...

Question

One mole of oxygen occupies a volume of $24 L$ at $20^{\circ} C$ and $1 a t m p r e s s s u r e$ . If the total momentum of the molecules in kgm/s is {10+x},then the value of x is

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Solution

Root mean square speed of the gas= $\sqrt{\frac{3 R T}{M_{0}}}$
$= \sqrt{\frac{3 \times R \times 293}{32 \times 10^{- 3}}} = 477.89 m / s$
Thus the total momentum of the molecules= $m v_{r m s} = 32 \times 10^{- 3} \times 477.89 k g m / s \approx 15 k g m / s$

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One mole of oxygen occupies a volume of 24L at 20∘C and 1atmpresssure. If the total momentum of the molecules in kgm/s is {10+x},then the value of x is

Root mean square speed of the gas=√3RTM0=√3×R×29332×10−3=477.89m/sThus the total momentum of the molecules=mvrms=32×10−3×477.89kgm/s≈15kgm/s

One mole of oxygen occupies a volume of $24 L$ at $20^{\circ} C$ and $1 a t m p r e s s s u r e$ . If the total momentum of the molecules in kgm/s is {10+x},then the value of x is

Root mean square speed of the gas= $\sqrt{\frac{3 R T}{M_{0}}}$
$= \sqrt{\frac{3 \times R \times 293}{32 \times 10^{- 3}}} = 477.89 m / s$
Thus the total momentum of the molecules= $m v_{r m s} = 32 \times 10^{- 3} \times 477.89 k g m / s \approx 15 k g m / s$