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?
- Anonymous1 decade agoFavorite Answer
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
so your new volume is .02108 L, or 21 mL.
- NorrieLv 71 decade ago
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.
- Anonymous4 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!
- 1 decade ago
STP is defined as 0C (= 273C) and 1atm (= 760mmHg)
Using the combined gas law (P1V1/T1 = P2V2/T2)
800mmHg * .022L / 300K = 760mmHg * xL / 273K
x = .021L = 21mLSource(s): I am an AP Chemistry student at Thomas Jefferson High School for Science and Technology.