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15 gram of nitrogen is enclosed in a ves...

15 gram of nitrogen is enclosed in a vessel at a temperature of 300 K.

A

Work done is 0 J

B

The amount of heat required to double the root mean square velocity of the molecules of the gas is `1.0016xx10^(4)J`

C

Change in internal energy of nitrogen gas is `2.03xx10^(4)J`, it root mean square velocity is doubled

D

All of these

Text Solution

Verified by Experts

The correct Answer is:
A, B
The average kinetic energy of each molecule of mass m is given by
`(1)/(2)mv_(rms)^(2)=(3)/(2)KT` (where K is Boltzman.s constant) or `T=propv_(rms)^(2)`
So if `v_(rms)` is doubled, the new temperature (T.) will be four times the initial temperature, i.e., `T.=4T`.
Change in internal energy of nitrogen gas is given by
`DeltaU=nC_(V)=(T.-T)`
where n is the number of moles of the gas and `C_(V)` is the molar specific heat at constant volume.
`therefore n=(m)/(M)=(15)/(28)` [M(molecular weight of nitrogen) = 28]
Also, `C_(V)=(5)/(2)R` [Nitrogen is diatomic]
where R is the universal gas constant.
`therefore DeltaU=((15)/(28))xx(5)/(2)R(4T-T)" "[because T.=4T]`
`=(15)/(28)xx(5)/(2)xx(8.31)((300xx4)-300)[because T=300K]`
`=1.0016xx10^(4)J`
Now, from first law of thermodynamics, we have
`Q=DeltaU+W`
Here, the gas is enclosed in a vessel
`therefore DeltaV=0`
Work done, `W=PxxDeltaV=0`
`therefore Q=1.0016xx10^(4)+0=1.0016xx10^(4)J`
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