Question

An ideal gas is kept in a container with a movable piston. A small hole is made on opposite wall of container as shown, so that gas leaks out of the container at a rate R (no. of molecules escaping the container per second) Now gas is heated as well as compressed by an external agent such that its temperature becomes 9 times and pressure becomes 4 times its initial value. Now, the rate of gas leaking out will be

Answer 1

Explanation:

Expression for work done is given by,

W=PΔV

Given that

P= Constant

And volume is doubled from initial position.

W=P(2V−V)

=PV

But, by Ideal gas equation PV=nRT

For 1 mole gas, PV=RT

=2RT−RT=RT.

⇒W=RT

Answer 2

Answer:

For an ideal gas,

PV = nRT

Since the gas leaked out with R and

R ∝ (no. of molecules leaving / vol of the container)

R ∝ (no. of initial molecules -  no. of final molecules) / volume

Let,

P₁ V₁ T₁ be initial values

P₂ V₂ T₂ be final values

P₁ V₁ / n₁ T₁ = P₂ V₂ / n₂ T₂

P₁ V₁ / n₁ T₁ = 4P₂ V₂ / n₂ 9T₂

Since,

R ∝ n₁ / V₁

P₁ / R*T₁ = 4/9 (P₁ / R₂ * T₁)

R₂ = 4 / 9 R

Therefore, the rate of gas leaking out will be R₂ = 4 / 9 R

To know more about pressure, refer: https://brainly.in/question/17070850

To know more about volume, refer: https://brainly.in/question/25282116

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