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For an ideal gas in a sealed container, if you double the pressure and double the volume, while keeping everything else the same, what has happened to the temperature?
A. The temperature is one-quarter of its original value.
B. The temperature is now four times its original value.
C. The temperature is equal to its original value.
D. The temperature is now double its original value.
E. The temperature is now half its original value.



Answer :

tutorconsortium272

For an ideal gas in a sealed container, if pressure and volume are made double, then : B.) temperature is now four times its original value.

What happens when pressure and volume are doubled?

Ideal gas is an hypothetical condition of gas state where molecules of the gas are spaced widely apart that is the inter molecular force of attraction is negligible. It obeys the equation,

PV = n RT

Here, P is the absolute pressure of the gas

V is Volume occupied by the gas

n is number of moles of gas

R is Gas constant

T is Absolute temperature of the gas

Ideal gas at two different states are given as:

((P1).(V1))/(T1) = ((P2).(V2))/(T2)

Given, P2 = 2*P1, V2 = 2*V1

So, ((P1).(V1))/(T1) = ((2*P1).(2*V1))/(T2)

1/T1 = 4/T2

T2 = 4 times of T1

Temperature of the gas becomes four times on doubling the absolute pressure and temperature of gas.

To know more about ideal gas, refer

https://brainly.com/question/27870704

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