Answered

If you double the volume of a constant amount of
gas at a constant temperature, what happens to
the pressure?
MISSED THIS? Read Section 11.4; Watch KCV 11.4, IWE 11.2
(a) The pressure doubles.
(b) The pressure quadruples.
(c) The pressure does not change.
(d) The pressure is one-half the initial pressure.

Verified Answer :

GinnyAnswer

Let's analyze the situation using Boyle's Law. Boyle's Law states that for a given amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.

Boyle's Law can be mathematically represented as:

\(\[ P_1 \times V_1 = P_2 \times V_2 \]\)

Where:
- \(\( P_1 \)\) is the initial pressure
- \(\( V_1 \)\) is the initial volume
- \(\( P_2 \)\) is the final pressure
- \(\( V_2 \)\) is the final volume

Now, let's consider the situation described in the problem:

1. Initial State:
- The gas has an initial pressure (\(\( P_1 \)\)).
- The gas has an initial volume (\(\( V_1 \)\)).

2. After Doubling the Volume:
- The final volume (\(\( V_2 \)\)) is twice the initial volume, so \(\( V_2 = 2 \times V_1 \)\).

We can plug this information into Boyle's Law equation:

\(\[ P_1 \times V_1 = P_2 \times (2 \times V_1) \]\)

To isolate \(\( P_2 \)\), divide both sides of the equation by \(\( 2 \times V_1 \)\):

\(\[ P_2 = \frac{P_1 \times V_1}{2 \times V_1} \]\)

The \(\( V_1 \)\) terms cancel out, leaving us with:

\(\[ P_2 = \frac{P_1}{2} \]\)

This means the final pressure (\(\( P_2 \)\)) is one-half of the initial pressure (\(\( P_1 \)\)).

So, the answer is:

(d) The pressure is one-half the initial pressure.

Related Questions