A 10.0-L balloon contains helium gas at a pressure of 655 mmHg. What is the new pressure, in mmHg, of the helium gas ateach of the following volumes. if there is no change in temperature and amount of gas?a. 20.0 Lb. 250 Lc. 1500 mL

The pressure at 20L volume has been 327.5 mm Hg, at 250 L volume has been 26.2 mm Hg, and at 1500 ml has been 4,366.6 mm Hg.

The ideal gas equation helps in the determination of the pressure of the gas at constant temperature and volume.

The ideal gas equation can be given by:

PV=nRT

When V and n are constant. The relationship between the initial and final can be given by:

P1V1 = P2V2

P1 and V1 are the initial pressure and temperature, and P2 and V2 are the final pressure and temperature.

Given, P1 = 655 mm Hg, V1 = 10 L

(a) V2 = 20 L

655 10 = P2 20

P2 = 327.5 mm Hg.

(b) V2 = 250 L

655 10 = P2 250

P2 = 26.2 mm Hg

(c) V2 = 1500 ml

V2 = 1.5 L

655 10 = P2 1.5

P2 = 4,366.6 mm Hg

For more information about the pressure of the gas, refer to the link:

link

a. 327.5 mm Hg, b. 26.2 mm Hg, c. 4,366.7 mm Hg

Explanation:

We can use the Boyle's law to find the new pressure.

P₁V₁ = P₂V₂

Now P₁ = 655 mm Hg

V₁ = 10 L

V₂ = 20 L

Plugin the values in the equation, we will get the new pressures as,

a.

b.

c.