52
1 point
If there are 10.0 Liters of a gas in a balloon and
the balloon expands to occupy a volume of 28.0
Liters with a final pressure of 4.00 atm, what was
the initial pressure? (The temperature is held
constant.)



Answer :

Sure, I'd be happy to help you solve this problem step-by-step!

Given:
- Initial volume of the gas, [tex]\( V_1 = 10.0 \)[/tex] liters
- Final volume of the gas, [tex]\( V_2 = 28.0 \)[/tex] liters
- Final pressure of the gas, [tex]\( P_2 = 4.00 \)[/tex] atm

We need to find the initial pressure, [tex]\( P_1 \)[/tex].

To solve this, we will use Boyle's Law, which states that for a given mass of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. Mathematically, Boyle's Law is expressed as:

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

From this equation, we can solve for the initial pressure [tex]\( P_1 \)[/tex]:

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

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

[tex]\[ P_1 = \frac{4.00 \, \text{atm} \times 28.0 \, \text{liters}}{10.0 \, \text{liters}} \][/tex]

First, perform the multiplication in the numerator:

[tex]\[ 4.00 \, \text{atm} \times 28.0 \, \text{liters} = 112.0 \, \text{atm}\cdot\text{liters} \][/tex]

Next, divide by the initial volume:

[tex]\[ P_1 = \frac{112.0 \, \text{atm}\cdot\text{liters}}{10.0 \, \text{liters}} \][/tex]

[tex]\[ P_1 = 11.2 \, \text{atm} \][/tex]

Therefore, the initial pressure was [tex]\( 11.2 \, \text{atm} \)[/tex].

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