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An ideal gas is initially at temperature T and volume V. Its volume is increased by `DeltaV` due to an increase in temperature `DeltaT,` pressure remaining constant. The quantity `delta=(DeltaV)/(VDeltaT)` varies with temperature asA. B. C. D. |
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Answer» Correct Answer - C c. From ideal gas equation `PV = RT` `P Delta T = R Delta T` Dividing Eq. (ii) by Eq. (i), we get `(Delta T)/(V) = (Delta T)/(T) implies (Delta V)/(V Delta T) = (1)/(T) = delta` (given) `:. Delta = (1)/(T)` So the graph between `delta` and `T` will be a rectangular hyperbola. |
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