Question

A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what will be the volume of the balloon if 0.20 moles of gas are added?

Answer 1

Answer:

V = 1.6 L

Explanation:

assuming ideal gas:

• PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 1.50 L

∴ n1 = 3.00 mol

∴ T1 = 25°C ≅ 298 K

⇒ P1 = (RT1n1)/(V1) = ((0.082 atm.L/K.mol)(298 K)(3.00 mol))/(1.50 L)

⇒ P1 = 48.872 atm

with pressure and temperature held constant:

∴ T2 = T1 = 298 K

∴ P2 = P1 = 48.872 atm

∴ n2 = 0.20 mol + 3.00 mol = 3.20 mol

⇒ V2 = (RT2n2)/P2

⇒ V2 = ((0.082 atm.L/K.mol)(298 K)(3.20 mol))/(48.872 atm)

⇒ V2 = 1.6 L

Answer 2

A balloon initially filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C, to which 0.20 moles of gas are added at constant pressure and temperature, will have a final volume of 1.60 L.

A balloon is filled to a volume of 1.50 L (V₁) with 3.00 moles of gas (n₁) at 25 °C. When 0.20 moles of gas are added, the final number of moles (n₂) will be:

$n_2 = 3.00 mol + 0.20 mol = 3.20 mol$

With pressure and temperature held constant, we can calculate the final volume of the balloon (V₂) using Avogadro's law, which states that there is a direct relationship between the number of gaseous moles and the volume.

$\frac{V_1}{n_1} = \frac{V_2}{n_2} \\V_2 = \frac{V_1 \times n_2 }{n_1} = \frac{1.50 L \times 3.20mol }{3.00mol} = 1.60 L$

A balloon initially filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C, to which 0.20 moles of gas are added at constant pressure and temperature, will have a final volume of 1.60 L.

You can learn more about Avogadro's law here: https://brainly.com/question/4133756