Question

An ideal gas is initially at temperature T and volume V. It's volume increases by DV due to an increase in temperature of DT, pressure remaining constant. The quantity m = $\frac{\Delta V}{V\Delta T}$ varies with temperature as –

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

Answer 2

V = KT ̃ DV = K DT ̃ $\frac{\Delta V}{\Delta T}$ = K …….(i)

̃ m = $\frac{\Delta V}{V\Delta T}$ ̃ m a $\frac{1}{V}$ ̃ m a $\frac{1}{T}$

So graph bet. m & T will be rectangular hyperbola.