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Now, let's solve the problem using Boyle's Law:

A gas exerts a pressure of 151 mm Hg in a container with an unknown volume. It exerts a pressure of 166 mm Hg when transferred to a 0.532 L container.

Given:
[tex]\[ P_1 = 151 \, \text{mm Hg} \][/tex]
[tex]\[ V_1 = ? \, \text{L} \][/tex]
[tex]\[ P_2 = 166 \, \text{mm Hg} \][/tex]
[tex]\[ V_2 = 0.532 \, \text{L} \][/tex]

Assume that the number of moles and the temperature remain constant. Find [tex]\( V_1 \)[/tex].



Answer :

GinnyAnswer
Alright, let's solve this problem step-by-step using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship when temperature and the number of moles are constant. Mathematically, this is expressed as:

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

Given data:
- Initial pressure, [tex]\( P_1 = 151 \text{ mm Hg} \)[/tex]
- Final pressure, [tex]\( P_2 = 166 \text{ mm Hg} \)[/tex]
- Final volume, [tex]\( V_2 = 0.532 \text{ L} \)[/tex]

We need to determine the initial volume, [tex]\( V_1 \)[/tex].

Rearranging the formula to solve for [tex]\( V_1 \)[/tex]:

[tex]\[ V_1 = \frac{P_2 \times V_2}{P_1} \][/tex]

Now let's substitute the given values into the equation:

[tex]\[ V_1 = \frac{166 \text{ mm Hg} \times 0.532 \text{ L}}{151 \text{ mm Hg}} \][/tex]

Carrying out the multiplication and division:

[tex]\[ V_1 = \frac{88.312 \text{ mm Hg} \cdot \text{L}}{151 \text{ mm Hg}} \][/tex]

[tex]\[ V_1 = 0.5848476821192053 \text{ L} \][/tex]

Thus, the initial volume [tex]\( V_1 \)[/tex] is approximately [tex]\( 0.5848 \text{ L} \)[/tex].

Therefore, the gas that initially exerts a pressure of 151 mm Hg in the original container has a volume of [tex]\( \boxed{0.5848 \text{ L}} \)[/tex].

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