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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Solution

$\begin{array}{l} Here, \\ RH=40% \\ V_{1} = 10 \times 10^{- 6} m^{3} \\ RH= \frac{V P}{S V P} = 0.4 \\ {Let SVP = P}_{o} \\ {Condensation occurs when VP = P}_{o} . \\ \Rightarrow P_{1} = 0.4 P_{o} \\ \Rightarrow P_{2} = P_{o} \\ Since the process is isothermal, applying Boyle's law we get \\ P_{1} V_{1} = P_{2} V_{2} \\ \Rightarrow V_{2} = \frac{P_{1} V_{1}}{P_{2}} \\ \Rightarrow V_{2} = \frac{0.4 P_{o} \times 10 \times 10^{- 6}}{P_{o}} \\ \Rightarrow V_{2} = 4.0 \times 10^{- 6} \\ \Rightarrow V_{2} = 4.0 {cm}^{3} \\ Thus water vapour condenses at volume 4 {.0 cm}^{3} . \end{array}$

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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.

Here,RH=40%V1=10×10−6m3RH=VPSVP=0.4Let SVP = PoCondensation occurs when VP = Po.⇒P1=0.4Po⇒P2=PoSince the process is isothermal, applying Boyle's law we getP1V1=P2V2⇒V2=P1V1P2⇒V2=0.4Po×10×10−6Po⇒V2=4.0×10−6⇒V2=4.0cm3Thus water vapour condenses at volume 4.0 cm3.

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.