Question #0c37e

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Question #0c37e

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Answer 1 · 2017-02-15T11:08:28

Given

$Total pressure of moisture and He P = 1.20 a t m$ $"Total pressure of moisture and He "P=1.20atm$

$Partial pressure of sat.vapour at 45^{\circ} C = (f) = 71.88 torr = \frac{71.88}{760} a t m \approx 0.095 a t m$ $"Partial pressure of sat.vapour at" 45^@C=(f)=71.88"torr"=71.88/760atm~~0.095atm$

$Temperature T = (45 + 273) = 328 K$ $"Temperature "T=(45+273)=328K$

$Pressure of He = p_{H e} = 1.20 - 0.95 = 1.105 a t m$ $"Pressure of He"=p_(He)=1.20-0.95=1.105atm$

$Mass of He = 0.586 g$ $"Mass of He"=0.586g$

$n_{H e} = \frac{0.586 g}{4 g /mol} = 0.1465 m o l$ $n_(He)=(0.586g)/(4g"/mol")=0.1465mol$

Now by ideal gas equation

$p_{H e} V_{H e} = n_{H e} R T$ $p_(He)V_(He)=n_(He)RT$

$V_{H e} = \frac{0.1465 m o l \times 0.082 L a t m K^{- 1} m o l^{- 1} \times 328 K}{1.105 a t m}$ $V_(He)=(0.1465molxx0.082LatmK^-1mol^-1xx328K)/(1.105atm)$

$\approx 3.57 L$ $~~3.57L$

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Question #0c37e

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