Question

Let P and V represent the pressure and volume of the Xe(g) in the container in diagram 3. If a piston is used to reduce the volume of the gas to V2 at a constant temperature, what is the new pressure in the container in terms of the original pressure, P ?

Answer 1

Answer:

2p

Explanation:

To solve this question, we can use Boyle's Law, which states that:

"For a fixed mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"

Mathematically:

$p\propto \frac{1}{V}$

where

p is the pressure of the gas

V is its volume

The equation can be rewritten as

$p_1 V_1 = p_2 V_2$

where in this problem we have:

$p_1 = p$ is the initial pressure of the Xe(g) gas

$V_1=V$ is the initial volume of the Xe(g) gas

$V_2=\frac{V}{2}$  is the final volume of the Xe(g) gas

Solving for p2, we find the final pressure of the gas:

$p_2=\frac{p_1 V_1}{V_2}=\frac{pV}{V/2}=2p$

So, the final pressure is twice the initial pressure.

Answer 2

Xe(g) is in a container at a certain pressure (P) and volume (V), If the volume is halved (V₂ = V/2), the pressure is doubled at a constant temperature. That is, P₂ = 2 P.

Let P and V represent the pressure and volume of the Xe(g) in a container. If a piston is used to reduce the volume of the gas to V/2 (V₂) at a constant temperature, we can calculate the new pressure (P₂) using Boyle's law, which establishes that there is an inverse relationship between the pressure and the volume of a gas.

$P \times V = P_2 \times V_2\\\\P_2 = \frac{P \times V}{V_2} = \frac{P \times V}{V/2} = 2P$

Xe(g) is in a container at a certain pressure (P) and volume (V), If the volume is halved (V₂ = V/2), the pressure is doubled at a constant temperature. That is, P₂ = 2 P.

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