Question

An ideal gas is initially at temperature T and volume V. Its volume is increased by $\Delta$V due to an increase in temperature $\Delta$T, pressure remaining constant. The quantity 5 = $\Delta$V / V$\Delta$T varies with temperature as

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

Answer 2

For an ideal gas, pV = nRT
For $\, \, \, \, \, \, \, \, \, \,$ p = constant
$\, \, \, \, \, \, \, \, \, \, p \Delta V =nR\Delta T$
$\therefore \, \, \, \, \, \, \, \, \frac{\Delta V}{\Delta T}=\frac{nR}{p}=\frac{nR}{\frac{nRT}{v}}=\frac{V}{T}$
$\therefore \, \, \, \, \, \, \, \, \, \, \frac{\Delta V}{V \Delta T}=\frac{1}{T}$
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.