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An ideal gas is initially at temperature \mathrm{T} and volume \mathrm{V}. It volume is increased by \Delta \mathrm{V} due to an increase in temperature \Delta \mathrm{T}, pressure remaining constant. The quantity \delta=\Delta \mathrm{V} / \mathrm{V} \Delta \mathrm{T} varies with temperature as :

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)

best_answer

For an ideal gas : \mathrm{PV}=\mathrm{nRT}

\mathrm{Since, \quad \mathrm{P}=\frac{n R T}{V} (given) = constant ( given)}
\mathrm{Therefore, \mathrm{P} \Delta \mathrm{N}=\mathrm{nR} \Delta \mathrm{T}}

\mathrm{\therefore \quad \frac{\Delta V}{D T}=\frac{n R}{P}=\frac{n R}{\frac{n R T}{V}}=\frac{V}{T}}


\mathrm{\therefore \quad \frac{\Delta V}{V \Delta T}=\frac{1}{T} \: or \: \delta=\frac{1}{T}}

Therefore, \delta is inversely proportional to temperature T. i.e., when T increases, \delta decreases and vice-versa.
Hence. \delta -T graph will be a rectangular hyperbola as shown in the above figure.

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Shailly goel

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