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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