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

Question

An ideal gas is initially at temperature T and volume V. Its volume is increased by ΔV due to an increase in temperature ΔT, pressure remaining constant. The quantity 5 = ΔV / VΔT varies with temperature as (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For an ideal gas, pV = nRT
For

p = constant

p Δ V = n R Δ T

∴ \frac{Δ V}{Δ T} = \frac{n R}{p} = \frac{n R}{\frac{n RT}{v}} = \frac{V}{T}

∴ \frac{Δ V}{V Δ T} = \frac{1}{T}

or

δ = \frac{1}{T}

Therefore,

δ

is inversely proportional to temperature T. i.e.
when T increases,

δ

decreases and vice-versa.
Hence,

δ -

T graph will be a rectangular hyperbola as shown in
the above figure.

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

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