What is the pressure of gas if 300 ml at 0.8 atm and 37degress C changes to 50 degrees C and 555ml?
Answer:
The Combined Gas Law.
P1 x V1 x T2 = P2 x V2 x T1
0.8atm x 300mL x 37°C (310K) = P2 X 555mL x 50°C (323).
P2 = (0.8 x 300 x 323) ÷ (555 x 310)
P2 = 77,520 ÷ 172,050
Final pressure (P2) = 0.45atm.
Assuming it's an ideal gas: 0.45 atm.
From the equation pV = nRT
we get nR = pV / T
Since n and R are constant we can get the combined gas' law
p1V1 / T1 = p2V2 /T2
p1 = 0.8 atm ; V1 = 0.300 L ; T1 = 310 K
p2 = ? ; V2 = 0.555 L ; T2 = 323 K
0.8 x 0.3 / 310 = p2 x 0.555 / 323
p2 = 0.450 atm
first of all, the units should be converted...
degree C into degrees K... 37 +273 =310 degree K & 50+273=323 degree K
300 ml = 0.3 L and 555 ml = 0.555 L
use the equation of state for ideal gases:
PV/T=P'V'/T'
(0.8)(0.3)/(310) = P' (0.555)/(323)
>>> P' = 0.45 atm
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