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Answer: 26.5 mm Hg
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:

where,
= initial pressure at
= ?
= final pressure at
= 100 mm Hg
= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature =
Now put all the given values in this formula, we get
![\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BP_1%7D%7B100%7D%29%3D%5Cfrac%7B28000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B299.5%7D-%5Cfrac%7B1%7D%7B267.9%7D%5D)



Thus the vapor pressure of
in mmHg at 26.5 ∘C is 26.5
Answer:
Yes
Explanation:
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The temperature of the gas sample is 813 K.
<u>Explanation:</u>
We have to use the ideal gas equation to find the temperature of the gas sample.
The ideal gas equation is PV = nRT
Pressure, P = 429 mm Hg = 0.56 atm
Volume, V = 560 mL = 0.56 L
R = gas constant = 0.08205 L atm mol⁻¹K⁻¹
Mass = 0.211 g
Molar mass of carbon di oxide = 44.01 g / mol
Moles, n = 
= 0.0047 mol
Now, we have to plugin the above values in the above equation, we will get the temperature as,

T = 
= 813 K
So the temperature of the gas sample is 813 K.