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//V DeltaT` varies with temperature as

Options

A.
B.
C.
D.

Answer

C

Explanation

For an ideal gas: `p_V = nRT`
For p = constant
`p Delta V = n R Delta T `
`:." "(Delta V)/(Delta T) = (nR)/p= (nR)/((nRT)/V)= V/T" ":." "(Delta V)/(V Delta T) = 1/T " or "delta = 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.