A student is performing an experiment using a resonance column and a tuning fork of frequency `244s^(-1)`. He is told that the air in the tube has been replaced by another gas (assuming that the air column remains filled with the gas). If the minimum height at which resonance occurs is `(0.350+- 0.005)m`, the gas in the tube is (Useful information : `sqrt(167RT) = 640J^(1//2)mol^(-1//2)`,
`sqrt(140RT) = 590J^(1//2)mol^(-1//2)`. The molar masses `M` in grams are given in the options. take the values of `sqrt((10)/(M))` for each gas as given there.)

Options

A. Neon(M=20,`sqrt((10)/(20))=(7)/(10)`)
B. Nitrogen(M=28,`sqrt((10)/(20))=(3)/(5)`)
C. Oxygen(M=32,`sqrt((10)/(32))=(9)/(19)`)
D. Argon(M=36,`sqrt((10)/(36))=(17)/(32)`)

Answer

D

Explanation

For minimum height, `(lambda)/4=e`
`(lambda)/4=0.356 pm 0.0005 or lambda=1.400 pm0.020`
`v=flambda=244(1.4000 pm0.020) implies v=(341.6pm4.88)m/8`
`v= sqrt((gammaRT)/(Mxx10^(-3)))=sqrt((100gammaRTxx10)/(M))`
For monoatomic `gamma=1.67, v=sqrt(167 RT) sqrt((10)/(M))`
For neon (monoatomic): `640xx7/10=448ms^(-1)`
For argon (monoatomic): `640xx17/32=340 ms^(-1)`
For diatomic `gamma =1.4: v=sqrt(140RT) sqrt((10)/(m))`
For nitogen (diatomic):`590xx3/5=384m s^(-1)`
For oxugen (diatomic): `590xx9/16=351.875 ms^(-1)`
Hence possible answer is (D).