A spherical glass container of unknown volume contains helium gas at 25°C and 1.950 atm. When a portion of the helium is withdrawn and adjusted to 1.00 atm at 23°C, it is found to have a volume of 1.55 cm3. The gas remaining in the first container shows a pressure of 1.710 atm. Calculate the volume of the spherical container.

in mL

answered earlier.

To find the volume of the spherical container, we can use the ideal gas law:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant
T = Temperature in Kelvin

First, let's convert the temperatures from Celsius to Kelvin:

T1 = 25°C + 273.15 = 298.15 K (initial temperature)
T2 = 23°C + 273.15 = 296.15 K (final temperature)

Next, let's calculate the number of moles of helium gas using the ideal gas law. We can assume both the initial and final states of the helium gas have the same number of moles (since only a portion was withdrawn):

PV = nRT

n = (PV) / (RT)

Let's calculate the number of moles of helium gas using the initial and final pressure and temperature:

n = (P1V1) / (R*T1)
n = (1.950 atm * V1) / (R * 298.15 K)

Similarly, we can calculate the number of moles of helium gas in the final state:

n = (P2V2) / (R*T2)
n = (1.00 atm * 1.55 cm^3) / (R * 296.15 K)

We can set these two equations equal to each other:

(1.950 atm * V1) / (R * 298.15 K) = (1.00 atm * 1.55 cm^3) / (R * 296.15 K)

Now, let's solve for V1, the initial volume of the spherical container:

V1 = (1.00 atm * 1.55 cm^3 * 298.15 K) / (1.950 atm * 296.15 K)

V1 = (1.55 cm^3 * 298.15 K) / 1.950

Finally, let's convert the volume from cm^3 to mL:

V1 = (1.55 cm^3 * 298.15 K) / 1.950 * 1 mL/1 cm^3

V1 = (1.55 * 298.15) / 1.950 mL

V1 ≈ 238.03 mL

Therefore, the volume of the spherical container is approximately 238.03 mL.