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Calculate the volume (mL) that the dry gas would occupy at STP?

I missed my teacher explaining this to the class, but here is the information I have:
lab temperature-22 degrees C
lab pressure: 775 mmHg
Gas volume 24mL
partial pressure of dry H2- 753.9 torr

STP correspond to 273 K (0° Celsiu) and 1 atm = (760 torr) pressure.
--torr and mmHg are same--

dry H2 ha 753.9 torr pressure instead of 760 torr, so Im using that for normal pressure. (Not entirely sure if I understand that part)

convert temperature to kelvin 22C + 273C = 295K

INITIAL (pressure * volume) / temperature = FINAL (pressure * volume) / temperature

P1V1 / T1 = P2V2 /T2

So lab conditions are left input side, and final STP is right side:

(775 torr* 24 ml) / 295K = (753.9 torr * "X" ml) / 273K

solve for "X" ml:

63.05 = 2.76* X
X = 22.8 ml

A sample of hydrojen ga wa collected over water that wa measured to be 21.0 C on a day when barometric pressure was 753.2 mm Hg. What would be the pressure of the dry hydrogen gas under laboratory condition? Hint: Use equation 7; PH2= P=-P

STP is standard temp and pressure = 25C & 1 atm (760 torr).
You should use PVT1=PVT2
Always change C to Kelvin, which is C+273.
Just plug and chug.

P1 775/760 = 1.02 atm, P2 = 1 atm
V1 = 24ml (0.024L), V2 unknown
T1 => -22+273= 251, T2=> 25+273= 298

Now its just math, and you are solving for V2.
1.02(0.024)(251) = 1(x)(298)
6.144 = 298x
6.144/298 = 0.0206 L (V2), x1000ml/L = 20.6ml