If we plot a graph between volume (L) and temperature $(^{\circ} C)$ by studying their variation for 2.0 g of certain ideal gas at 1 bar pressure, the graph obtained is a straight line which is

Parallel to the temperature axis

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Parallel to the volume axis

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Meets the temperature axis where

T = 0, V = 0

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Meets the temperature axis where

V = 0; T = - 273.15

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Solution

The correct option is D Meets the temperature axis where $V = 0; T = - 273.15$
Given Weight =2g
Here Pressure and n (no af moles) are constant
Applying ideal gas equation.
$\begin{matrix} P V = N R T V = \frac{n R}{P} \cdot T V = K . T \end{matrix}$ V is zero at 0 K=-273.15C
The straight line meets the temperature axis.
where V=0 , T=-273.15C.

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Similar questions

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P (i . e ., \frac{1}{P})

for an ideal gas at constant temperature T, the curve so obtained is _______.

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Q. Arrange the relevant points in the proper sequence for explaining why Kelvin scale is preferred to Celsius scale in the study of gases.
(a)

- 273^{o}

C is the least possible temperature.
(b) A graph of volume vs temperature is a straight line passing through origin.
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(^{0} C)

is a straight line which intersects volume axis at some point
(d) Extrapolation of straight line touches the volume axis at

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(e)

273^{o}

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0 K

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If we plot a graph between volume (L) and temperature (∘C) by studying their variation for 2.0 g of certain ideal gas at 1 bar pressure, the graph obtained is a straight line which is

The correct option is D Meets the temperature axis where V=0;T=−273.15Given Weight =2gHere Pressure and n (no af moles) are constantApplying ideal gas equation.PV=NRTV=nRP⋅TV=K.TV is zero at 0 K=-273.15CThe straight line meets the temperature axis.where V=0 , T=-273.15C.

If we plot a graph between volume (L) and temperature $(^{\circ} C)$ by studying their variation for 2.0 g of certain ideal gas at 1 bar pressure, the graph obtained is a straight line which is