A flexible vessel contains 43L of gas where the pressure is 1.3 atm. What will the volume be when the pressure is 0.61 atm, the temperature remaining constant?

The volume will be 91.64 L when the pressure is 0.61 atm

From Boyle's law equation, we have that

P₁V₁ = P₂V₂

Where P₁ is the initial pressure

V₁ is the initial volume

P₂ is the final pressure

and V₂ is the final volume

From the given information

P₁ = 1.3 atm

V₁ = 43 L

P₂ = 0.61 atm

V₂ = ?

Putting the parameters into the formula, we get

1.3 × 43 = 0.61 × V₂

∴

V₂ = 91.64 L

Hence, when the pressure is 0.61 atm, the volume will be 91.64 L

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91.64L

Explanation:

Step 1:

Data obtained from the question.

This includes the following:

Initial volume (V1) = 43L

Initial pressure (P1) = 1.3 atm

Final volume (V2) =..?

Final pressure (P2) = 0.61 atm

Step 2:

Determination of the final volume.

To solve for the final volume, we'll apply the Boyle's law equation since the temperature is constant. The final volume is obtained as follow:

P1V1 = P2V2

1.3 x 43 = 0.61 x V2

Divide both side by 0.61

V2 = (1.3 x 43)/0.61

V2 = 91.64L

Therefore, the new volume is 91.64L

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :