Pure water vapour is trapped in a vessel of volume 10 $c m^{3}$ . The relative humidity is 40 %. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

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Solution

$R_{H} = \frac{V P}{S V P}$

The point where the vapour start condensing,

VP= SVP

We know $P_{1} V_{1} = P_{2} V_{2}$

$R_{H} \times S V P \times 10 = S V P \times V_{2}$

$\Rightarrow V_{2} = 10 R_{H}$

= 10 $\times$ 0.4

= 4 cm

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Q. Pure water vapour is trapped in a vessel of volume 10 cm³. The relative humidity is 40%. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

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Pure water vapour is trapped in a vessel of volume 10 cm3. The relative humidity is 40 %. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

RH=VPSVP The point where the vapour start condensing, VP= SVP We know P1V1=P2V2 RH×SVP×10=SVP×V2 ⇒V2=10RH = 10 × 0.4 = 4 cm

Pure water vapour is trapped in a vessel of volume 10 $c m^{3}$ . The relative humidity is 40 %. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

$R_{H} = \frac{V P}{S V P}$

The point where the vapour start condensing,

VP= SVP

We know $P_{1} V_{1} = P_{2} V_{2}$

$R_{H} \times S V P \times 10 = S V P \times V_{2}$

$\Rightarrow V_{2} = 10 R_{H}$

= 10 $\times$ 0.4

= 4 cm