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15 gram of nitrogen is enclosed in a vessel at a temperature of 300 K. |
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Answer» Work done is 0 J `(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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