The volume of a gas is 5.80L, measured at 1.00atm. What is the pressure of the gas in mmHg if the volume is changed to 9.65L ( temperature is kept constant)

Boyle's law is defined as the gas law, in which the pressure and volume of a gas have an inverse relationship. It determines how the pressure of a gas tends to decrease as the volume increases.

The new pressure of the gas is 456.76 mmHg.

Given that:

The initial volume of the gas = 5.80 L

The initial pressure of the gas = 1.00 atm

The volume of the gas increases = 9.65 L

From Boyle's law, the relation between pressure and volume at constant temperature can be explained as:

P₁V₁ = P₂V₂

where,

P₁ = initial pressure of the gas = 1.00 atm

V₁ =  initial volume of the gas = 5.80 L

P₂ = ?

V₂ =  increased volume of the gas = 9.65 L

Substituting the values, we get:

Now, converting atm into mmHg, we get:

1 atm = 760 mmHg

0.6010 atm = 456.76 mmHg.

Therefore, the new pressure is 456.76 mmHg.

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The new pressure is 456.76 mmHg

Explanation:

Step 1: Data given

The initial volume of the gas = 5.80 L

The initial pressure of the gas = 1.00 atm

The volume of the gas increases to 9.65 L

The temperature is kept constant

Step 2: Calculate the new pressure

P1*V1 = P2¨V2

⇒with P1 = the initial pressure of the gas = 1.00 atm

⇒with V1 = the initial volume of the gas = 5.80 L

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the increased volume of the gas = 9.65 L

1.00 atm * 5.80 L = P2 * 9.65 L

P2 = (5.80 / 9.65)

P2 = 0.6010 atm

Step 3: Convert pressure from atm to mm Hg

0.6010 atm * 760 mmHg = 456.76 mmHg

The new pressure is 456.76 mmHg

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