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प्रश्न
Can a process on an ideal gas be both adiabatic and isothermal?
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उत्तर
According to the first law of thermodynamics, change in internal energy, ΔU is equal to the difference between heat supplied to the gas, Δ Q and the work done on the gas, ΔW,
such that ΔQ = ΔU +ΔW . In an adiabatic process, ΔQ =0 and in an isothermal process, change in temperature, Δ T =0. Therefore,
ΔQ = ΔU + ΔW
⇒ ΔQ = nCvΔT +ΔW
⇒ 0 =nCv(0) + Δ W
⇒ Δ W = 0 ,
if the process is adiabatic as well as isothermal, no work will be done. So, a process on an ideal gas cannot be both adiabatic and isothermal.
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संबंधित प्रश्न
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
| Substance | Atomic Mass (u) | Density (103 Kg m-3) |
| Carbon (diamond) | 12.01 | 2.22 |
| Gold | 197.00 | 19.32 |
| Nitrogen (liquid) | 14.01 | 1.00 |
| Lithium | 6.94 | 0.53 |
| Fluorine (liquid) | 19.00 | 1.14 |
[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].
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