An ideal gas is initially at temperature T and volume V- Its volume is increased by - V due to an increase in temperature - T- pressure remaining constant- The quantity - VV - T varies with temperature asTemperature KTemperature K Temperature KV

The correct option is D For a given mass of gas at constant pressure, VT=Constant ∴V+∆VT+∆T=VT nbsp; or nbsp; VT+V∆T=VT+T∆V or nbsp; V∆T=T∆V or 1T=∆VV∆T ...

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Byju's Answer

Standard XII

Chemistry

Molar and Specific Heat Capacity

An ideal gas ...

Question

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 $δ = \frac{Δ V}{V Δ T}$ varies with temperature as

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Solution

The correct option is C
For a given mass of gas at constant pressure,
$\frac{V}{T} = C o n s t a n t$
$∴ \frac{V + Δ V}{T + Δ T} = \frac{V}{T}$
or
$V T + V Δ T = V T + T Δ V$
or
$V Δ T = T Δ V$
or
$\frac{1}{T} = \frac{Δ V}{V Δ T}$
or
$\frac{1}{T} = δ$
or
$δ T = 1$
This equation represents a rectangular hyperbola of the form $x y =1$ .
Hence the $δ - T$ variation is represented by graph given in (c).

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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 δ=ΔVVΔT varies with temperature as

The correct option is C For a given mass of gas at constant pressure, VT=Constant ∴V+ΔVT+ΔT=VT or VT+VΔT=VT+TΔV or VΔT=TΔV or 1T=ΔVVΔT or 1T=δ or δT=1 This equation represents a rectangular hyperbola of the form xy=1. Hence the δ−T variation is represented by graph given in (c).

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 $δ = \frac{Δ V}{V Δ T}$ varies with temperature as