How many moles of a gas will 700 m hold at -100 oC and 86 atm?
1 answer:
0.23 moles of a gas will 700 m hold at -100 oC and 86 atm.
Explanation:
given that:
volume = 700 ml OR 0.7 Litre (for converting into litre it is divided with 1000 as 1 litre = 1000ml)
temperature = -100 degrees OR 173.15 K
pressure = 86 atm
R = 0.082057 L atm mol-1K-1
n=?
It is known that when Volume is taken in litres, pressure in atm, and temperature in Kelvin then R= 0.082057 L atm mol-1K-1,
From the formula for ideal gas:
PV = nRT
putting the values in the formula:
n =
=
= 0.23 moles of the gas will be required.
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Answer: 77.4 mL
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is:
where,
= initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg
= final pressure of dry gas at STP = 760 mm Hg
= initial volume of dry gas = 85.0 mL
= final volume of dry gas at STP = ?
= initial temperature of dry gas =
= final temperature of dry gas at STP =
Now put all the given values in the above equation, we get the final volume of wet gas at STP
Volume of dry gas at STP is 77.4 mL.