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.



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:

[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]

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

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

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

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

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

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

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

[tex]\[ P_2 = \frac{P_1 \times V_1}{2 \times V_1} \][/tex]

The [tex]\( V_1 \)[/tex] terms cancel out, leaving us with:

[tex]\[ P_2 = \frac{P_1}{2} \][/tex]

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

So, the answer is:

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

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