If the thermometer reads 22 c and this corresponds to 19.8mmhg of water pressure , what is the pressure of a pure gas collected over water if the total pressure is 760 mmhg?next calculate the pressure of the gas and lets assume its hydrogen gas in atm.
pressure of pure gas=760-19.8mmhg
=740.2mmhg. That would be my P2
P1*V1=P2*V2
v2=P1*V1/p2
V2=19.8mmhg*1atm/740.2mmhg
Would my answer be ==0.02674.
Any help will be appreciated.
Your first calculation of P = 740.2 mm pressure for the pure (dry) gas is correct. The last part involving P1V1 = P2V2 isn't and I don't think you can calculate a volume of the H2 gas because you list no volumes in the problem. The numbers you have substituted are pressure values.
To find the pressure of the pure gas collected over water, you would subtract the partial pressure of water vapor (19.8 mmHg) from the total pressure (760 mmHg). This gives you the pressure of the pure gas:
Pressure of pure gas = Total pressure - Partial pressure of water vapor
= 760 mmHg - 19.8 mmHg
= 740.2 mmHg
So, you correctly calculated the pressure of the pure gas as 740.2 mmHg (which is your P2).
To convert this pressure from mmHg to atm, you can divide by the conversion factor of 760 mmHg = 1 atm:
Pressure of the gas (in atm) = Pressure of the gas (in mmHg) / 760 mmHg/atm
= 740.2 mmHg / 760 mmHg/atm
≈ 0.97318 atm
Therefore, the pressure of the pure gas, assuming it is hydrogen gas, would be approximately 0.97318 atm.