if both pressure and temperature are doubled the volume will?
give reason
Answers
Answer:
If both the pressure and volume of a given sample of an ideal gas are doubled, what happens to the temperature of the gas in Kelvins?
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Piriya
Answered 1 year ago
An ideal gas is an hypothetical condition of gas state where the molecules of the gas are spaced widely apart i.e. the inter molecular force of attraction is so negligible. It obeys the following equation,
PV = nRT
P = Absolute pressure of the gas (Pa)
V = Volume occupied by the gas (m3)
n = number of moles of gas (moles)
R =Gas constant (J/kg.K)
T = Absolute temperature of the gas (K)
Any real gas approaches an ideal state at high temperatures and low pressure.
Average kinetic energy of the molecules are more at higher temperatures, that makes them travel larger distances, hence to be placed very much apart.
Ideal gas at two different states are represented by,
((P1).(V1))/(T1) = ((P2).(V2))/(T2)
As per your question, P2 = 2*P1, V2 = 2*V1
So the equation becomes,
((P1).(V1))/(T1) = ((2*P1).(2*V1))/(T2)
Cancelling P1 & V1 on both sides gives,
1/T1 = 4/T2
Rearranging the equation,
T2 = 4 times of T1
So the temperature of the gas quadruples on doubling the absolute pressure and temperature of the gas.
Answer:
At this point, the volume of the gas is halved. If the pressure on the piston is again doubled, the volume of gas decreases to one-fourth of its original volume.
Explanation:
The ideal gas law is the equation of the state of a hypothetical ideal gas. It is a good approximation of the behaviour of many gases under many conditions
PV=nRT,
where:
P is the pressure of the gas,
V is the volume of the gas,
n is the amount of substance of gas (in moles),
R is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant,
T is the absolute temperature of the gas
P 0 V 0 =nRT 0
When pressure and volume doubled,⇒2P 0×2V 0⇒4P 0V 0=4×nRT 0
=nR(4T 0 )⇒T=4T 0
Temperature is increased by 4 times.
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