# Standard Temperature and Pressure?

Im studying for a chem test and i dont understand STP, can you explain how to do this problem:

At 27 C and 800 mmHg pressure, a quantity of hydrogen has a volume of 22 mL. what would the volume be under standard conditions?

Relevance
• Anonymous

at STP, Pressure is 1atm, and Temperature is 273K. ALWAYS.

use (P)(V) / (t)= (P1)(V1) / (T1).

convert 27C to K. 27+273 = 300K.

convert 800mmHG to atm. 800/760=1.053atm.

convert ml to L. 22mL= .022L.

insert into the equation: (1.053atm)(.022)/300K = (1atm)(x L)/(273K)

solve for x, which is your volume.

(1.053atm)(.022)/300K = (1atm)(x L)/(273K)

==0.023166/300 = x/273

==0.00007722=x/273

x=.02108L

so your new volume is .02108 L, or 21 mL.

• Norrie
Lv 7

Standard conditions for gases are taken as 273 Kelvin (0°C) and 1.0 atmosphere of Pressure taken at sea level.

1.0 atm. is also equal to 760mmHg (or 760 torr); and, 101.325 kPa. (In Imperial, it's 14.7 psia).

For your question, the Combined Gas Law can be used...

(P1 x V1 x T2 = P2 x V2 x T1).

Pressure conversion to atmospheres and mL conversion to Litres, is not necessary.

P1 = 800mmHg; V1 = 22mL; T2 = 273K.

P2 = 760mmHg; V2 = ? and T1 = (27°C + 273) = 300K

Plug these values into the calculation.

800mmHg x 22mL x 273K = 760mmHg x V2 x 300K.

V2 (new volume) = (800 x 22 x 273) / (760 x 300).

= 4,804,800 / 228,000 = 21.07mL final volume at STP.

Source(s): Senior Technical Instructor.
• Anonymous
4 years ago

-273 C is 0 ok, which isn't an pretty popular temperature. no rigidity is likewise no longer an pretty popular rigidity. by potential of technique of removing, it is going to be D. i in my opinion did no longer understand that 0 C became the popular temp; i presumed it became 25 C. You learn something every day!