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

`\boxed {\boxed {\sf 4.6 \ atm}}`

Step-by-step 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} = (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.

Answer 2

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

Using boyles law

`\boxed{\sf v\propto (1)/(p)}`

`\\ \sf\longmapsto P_1V_1=P_2V_2`

`\\ \sf\longmapsto P_2=(P_1V_1)/(V_2)`

`\\ \sf\longmapsto P_2=(2.8* 1.5)/(0.92)`

`\\ \sf\longmapsto P_2=(4.2)/(0.92)`

`\\ \sf\longmapsto P_2=4.56atm`

`\\ \sf\longmapsto P_2\approx 4.6atm`