Question

At a constant temperature, a sample of gas occupies 1.5 L at a pressure of 2.8 ATM. What will be the pressure of this sample, in atmospheres, if the new volume is 0.92 L?

Answer 1

• V1=1.5L
• V2=0.92L
• P1=2.8atm
• P2=?

Using boyles law

$\boxed{\sf v\propto \dfrac{1}{p}}$

$\\ \sf\longmapsto P_1V_1=P_2V_2$

$\\ \sf\longmapsto P_2=\dfrac{P_1V_1}{V_2}$

$\\ \sf\longmapsto P_2=\dfrac{2.8\times 1.5}{0.92}$

$\\ \sf\longmapsto P_2=\dfrac{4.2}{0.92}$

$\\ \sf\longmapsto P_2=4.56atm$

$\\ \sf\longmapsto P_2\approx 4.6atm$

Answer 2

Answer:

$\boxed {\boxed {\sf 4.6 \ atm}}$

Explanation:

We are asked to find the new pressure given a change in volume. 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:

$P_1V_1= P_2V_2$

Initially, the gas occupies 1.5 liters at a pressure of 2.8 atmospheres.

$1.5 \ L * 2.8 \ atm = P_2V_2$

The volume is changed to 0.92 liters, but the pressure is unknown.

$1.5 \ L * 2.8 \ atm = P_2* 0.92 \ L$

We are solving for the final pressure, so we must isolate the variable P₂. It is being multiplied by 0.92 liters. The inverse operation of multiplication is division, so we divide both sides by 0.92 L.

$\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L} = \frac{P_2* 0.92 \ L}{0.92 \ L}$

$\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L}= P_2$

The units of liters cancel each other out.

$\frac {1.5 * 2.8 \ atm}{0.92 }=P_2$

$\frac {4.2}{0.92} \ atm= P_2$

$4.565217391 \ atm = P_2$

The original measurements of pressure and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 6 in the hundredth place tells us to round the 5 up to a 6.

$4.6 \ atm \approx P_2$

The pressure is approximately 4.6 atmospheres.