One mole of oxygen is at a pressure of 6 atm and a temperature of 27 degrees cenlsius. a. If the gas is heated so that both the pressure and volume are doubled, what is the final temperature.?b. Ifg the gas is heated at constant volume until the pressure triples, what is the final temperature?
2007-01-02 15:28:51 · 5 answers · asked by Kitana 2 in Science & Mathematics ➔ Physics
use the gas equation pv/T = constant
so, since pressure and volume doubles, clearly temperature becomes four times its original value. Now note that in this formula, T shud be in kelvins. Initial T = 300 K. (add 273). So, new T = 1200 K = 927 degrees celsius.
For second part, v remains same and p triples. so T triples. so new T = 900 K = 627 degrees celsius
2007-01-02 15:40:37 · answer #1 · answered by ? 3 · 1⤊ 0⤋
a. 27 C = 300 K. To double both the pressure and volume would require that the temperature be quadrupled, to 1200 K.
b. Same principle, except we are just tripling the temperature, to 900 K.
2007-01-02 23:37:10 · answer #2 · answered by Anonymous · 1⤊ 0⤋
Too bad, Niks. You have to quadruple the Kelvin temperature, not the Celcius one.
2007-01-02 23:43:46 · answer #3 · answered by ? 6 · 0⤊ 0⤋
pv=rt
6*v=R*27
a)
12*2V=R*t
DIVIDING
4*27 = T
T= 107 Celcius
b)
6*3*v=R*T
Dividing
T=81
2007-01-02 23:36:55 · answer #4 · answered by Niks 3 · 0⤊ 1⤋
PV / T is a constant.
(P2/P1) x (V2/V1) x (T1/T2) = 1.
It is given (P2/P1) and (V2/V1) = 2.
2 x 2 x (T1/T2) = 1
(T1/T2) = 1/4 or (T2/T1) = 4
T2 = 4 T1.
T2 = 4 (273 +27)
= 1200K
Or 927C.
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(P2/P1) x (V2/V1) x (T1/T2) = 1.
It is given P2/P1 = 3 and V2/V1= 1.
3 x (T1/T2) = 1.
T1/T2 = 1/3 or T2/T1 = 3
T2 = 3 T1 = 3 ((273 +27)
= 900 K
Or 627C.
2007-01-03 06:26:36 · answer #5 · answered by Pearlsawme 7 · 0⤊ 0⤋