An ideal gas at 20 degrees centigrade has a volume of 12 liters and a pressure of 15 atmospheres. Its temperature is raised to 3
1 answer:
P 15 atm
V₁ = 12x103 m3 -3 3 V₁ = 12x10 m" T₁ = 25°c T= 28815 K
P₁ = 1.52x10 Pa
Now, Finel quantities, T₂ = 35°C = 308.15 °K
V2= 851 = -3 3
Now
by ideal gas equation
P2Y2 T2
= PV, T2
1152 X106 x 12x16 3 x 308115 8,5×103 x 288,15 = Pa
=
21294×10 Pa
P₂ = 21264 Atm
final
pressure will be
21264 Adm.
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Answer: Concentration of in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of = 0.729 M
The given balanced equilibrium reaction is,
Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
Now put all the given values in this expression, we get :
concentration of in the equilibrium mixture =
Thus concentration of in the equilibrium mixture is 0.31 M