Question

A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is __________ atm. A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is __________ atm. 1.5 7.5 0.67 3.3 15

Answer 1

Answer:

The final pressure will be 0.67 atm.

Explanation:

Using Boyle's law  

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,  

V₁ = 10 L

V₂ = 15 L

P₁ = 1.0 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${1.0}\times {10}={P_2}\times {15} atm$

${P_2}=\frac{{1.0}\times {10}}{15} atm$

${P_2}=0.67\ atm$

The final pressure will be 0.67 atm.

Answer 2

Answer:

A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.

Explanation:

Boyle's law states that in constant temperature the  variation volume of gas is inversely proportional to the applied pressure.

The formula is,

$\rm \bold{ P_1\times V_1= P_2\times V_2}$

Where,

$\rm \bold P_1$ is initial pressure = 1 atm

$\rm \bold P_2$ is final pressure =  ? (Not Known)

$\rm \bold V_1$ is initial volume  = 10 L

$\rm \bold V_2$ is final volume = 15 L

Now put the values in the formula,

$\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67$

Therefore, the answer is 0.67 atm.

For more details, you can refer to the link:

https://brainly.com/question/4109864?referrer=searchResults