Question

At 1.70 atm , a sample of gas takes up 4.25 L. If the pressure in the gas is increased to 2.40 atm. what the new volume be of the gas. Temperature is constant. ​

Answer 1

Given:

• Initial pressure(p1) = 1.70 atm.

• Initial volume(v1)= 4.25 L.

• Final pressure(p2) = 2.40 atm.

• Temperature is constant.

To find:

The final volume of the gas(v2).

Solution:

Initial pressure(p1) = 1.70 atm.

Initial volume(v1)= 4.25 L.

Final pressure(p2) = 2.40 atm.

Temperature is constant.

Final volume(v2) = ?

From the formula we get that,

p1v1 = p2v2 [At constant temperature.]

⇒ (1.70×4.25) = (2.40×v2)

⇒ (2.40×v2) = (1.70×4.25)

⇒ (2.40×v2) = 7.225

⇒ v2 = (7.225÷2.40)

⇒ v2 = 2.89 L

∴ The final volume of the gas will be 2.89 L.

Answer:

             The final volume of the gas will be 2.89 L.

Answer 2

Answer:

The new volume of gas is 3.01L

Explanation:

Given,

Initial Pressure =1.70atm

Final Pressure=2.40atm

Initial Volume=4.25L

Final Volume=?

Since temperature is constant we can use relation of Boyle's Law i.e,(Pressure is inversely proportional to volume)

$P _{1}V _{1} = P_{2} V_{2} = > V_{2} = \frac{P _{1}V _{1}}{P_{2}}$

Therefore by substituting the given values we get the final unknown volume as

$V _{2} = \frac{1.70 \times 4.25}{2.40} = 3.01lit$

Therefore new volume of gas is 3.01L

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