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The kinetic energy of translation of 20g...

The kinetic energy of translation of 20gm of oxygen at `47^(@)C` is (molecular wt. of oxygen is 32 gm/mol and R=8.3 J/mol/K)

A

2490 ergs

B

2490 J

C

249 J

D

960 J

Text Solution

Verified by Experts

The correct Answer is:
B
For 1 gram molecule of a gas pV = RT
and for 1 gram of a gas PV = rT = `R/32T`
`therefore` For 20 gram, PV = 20rT = `20xxR/32xxT`
`thereforeK.E. = 3/2(20rT)=3/2xx20xx8.3/32xx(273+47)`
`=3/2xx20xx8.3/32xx320=2490J`
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