Question

A container is filled to a volume of 55.2 L at 61 °C. While keeping the

Answer 1

Answer:

$\boxed {\boxed {\sf 4.45 \ atmospheres}}$

Explanation:

We are asked to find the pressure given a change in volume. The temperature remains constant, so we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:

$P_1 V_1= P_2V_2$

The initial pressure is unknown, but the volume starts at 55.2 liters.

$P_1 * 55.2 \ L = P_2V_2$

The volume is reduced to 28.8 liters and the pressure is 8.53 atmospheres.

$P_1 * 55.2 \ L = 8.53 \ atm * 28.8 \ L$

We are solving for the initial pressure, so we must isolate the variable P₁. It is being multiplied by 55.2 liters. The inverse operation of multiplication is division, so we divide both sides of the equation by 55.2 L.

$\frac {P_1 * 55.2 \ L }{55.2 \ L}= \frac{8.53 \ atm * 28.8 \ L}{55.2 \ L}$

$P_1= \frac{8.53 \ atm * 28.8 \ L}{55.2 \ L}$

The units of liters (L) cancel.

$P_1= \frac{8.53 \ atm * 28.8 }{55.2}$

$P_1=\frac{245.664 }{55.2 } \ atm$

$P_1 = 4.45043478261 \ atm$

The original measurements of volume and pressure have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredths place. The 0 in the thousandths place tells us to leave the 5.

$P_1 \approx 4.45 \ atm$

The initial pressure inside the container is approximately 4.45 atmospheres.

Answer 2

Answer:

4.45 atm

Explanation:

Applying,

PV = P'V'............ Equation 1

Where P = Initial pressure of the container, V = Initial volume of the container, P' = Final pressure of the container, V' = Final volume of the container.

make P the subject of the equation

P = P'V'/V........... Equation 2

From the question,

Given: V = 55.2 L, P' = 8.53 atm, V' = 28.8 L

Substitute these values into equation 2

P = (8.53×28.8)/55.2

P = 4.45 atm