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 : The correct option is, pressure.
Explanation :
The ideal gas equation is,
where,
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
T = temperature of the gas
R = gas constant
The value of 'R' has several different values which are :
That means, the value of 'R' is different due the change in the pressure value and all the variables (temperature, volume and moles) are constant.
Hence, the correct option is, pressure.