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10 days ago

13

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?

2 answers:

VMariaS [2.9K]10 days ago

8 0

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

Using Boyle's law

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

$\\ \sf\longmapsto P_1V_1=P_2V_2$

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

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

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

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

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

KiRa [2.9K]10 days ago

7 0

Answer:

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

Explanation:

In this scenario, we need to determine the new pressure following a volume change, utilizing Boyle's Law, which states that gas volume is inversely related to pressure. The governing equation is:

$P_1V_1= P_2V_2$

Initially, the gas is at a volume of 1.5 liters with a pressure of 2.8 atmospheres.

$1.5 \ L * 2.8 \ atm = P_2V_2$

This volume decreases to 0.92 liters, but the new pressure is unknown.

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

To solve for the final pressure, we isolate P₂. The equation involves multiplication by 0.92 liters, and we counter this via division on both sides.

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

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

This allows the liters to cancel out.

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

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

Considering that the initial pressure and volume both hold 2 significant figures, our result must align with that. Thus, the number derived must be adjusted to the tenths place, where a rounding is indicated. $4.565217391 \ atm = P_2$

The final pressure value is close to 4.6 atmospheres.

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