A spherical glass container of unknown volume contains helium gas at 25C and 1.960 atm. When a portion of the helium is withdrawn and adjusted to 1.00 atm at 25C, it is found to have a volume of 1.75 cm^3. The fas remaining in the first container shows a pressure of 1.710 atm. Calculate the volume of the spherical container.

Use PV = nRT to calculate n for the gas removed.
Then substitute this n in PV = nRT for the shperical container using P = 1.960 - 1.710 and solve for V.
Post your work if you get stuck.

1 atm X 1.75 cm^3 = n X(.08206)X 298.15K

n = 0.0715?
For the container.

(1.960 - 1.710 atm) X Volume = 0.0715 X (0.08206) X 298.15K.

Volume = 0.1429 cm^3?

That small?

the stepsare correct but the arthimetic is wrong somewhere in between. You should get 7 ml

agree with dmoney, the n value(moles) does not equal 0.0715, its 7.156 ^-3 which would make the answer 6.9938... or 7mL. Do not convert from mL to L in between, either do it at the beginning or the end.

To calculate the volume of the spherical container, let's solve the equation using the given values:

1 atm * 1.75 cm^3 = n * (0.08206 L·atm/(mol·K)) * 298.15K

Simplifying the equation:

n = (1.75 cm^3) / (0.08206 L·atm/(mol·K) * 298.15K)
n ≈ 0.0715 mol

Now, substitute the value of n into the equation for the spherical container:

(1.960 atm - 1.710 atm) * Volume = 0.0715 mol * (0.08206 L·atm/(mol·K)) * 298.15K

Simplifying the equation:

Volume = (0.0715 mol * (0.08206 L·atm/(mol·K)) * 298.15K) / (1.960 atm - 1.710 atm)
Volume ≈ 0.1429 L

Converting from liters to cm^3:

Volume ≈ 0.1429 L * (1000 cm^3/1 L)
Volume ≈ 142.9 cm^3

So, the volume of the spherical container is approximately 142.9 cm^3.

To calculate the volume of the spherical container, we need to use the ideal gas law equation PV = nRT.

First, we need to calculate the moles of gas removed from the container using the given information. The pressure of the withdrawn portion of helium is adjusted to 1.00 atm. Therefore, we can use PV = nRT to solve for n, the number of moles of gas.

Using the values given:
P = 1.00 atm
V = 1.75 cm^3
R = 0.08206 L·atm/(mol·K)
T = 25 + 273.15 K (convert Celsius to Kelvin)

1.00 atm * 1.75 cm^3 = n * 0.08206 L·atm/(mol·K) * 298.15 K

Simplifying the equation:
1.75 cm^3 = 24.5397 * n

Solving for n:
n = 0.0715 moles (rounded to four decimal places)

Now, we have the number of moles of gas removed from the container. We can use PV = nRT again to calculate the volume of the spherical container.

The pressure remaining in the container is 1.710 atm (1.960 atm - 1.710 atm). Substituting the values into the equation:

(1.960 - 1.710 atm) * Volume = 0.0715 mol * (0.08206 L·atm/(mol·K)) * 298.15 K

Simplifying the equation:
0.25 atm * Volume = 1.424 L·atm

Solving for Volume:
Volume = 1.424 L·atm / 0.25 atm = 5.696 L

Finally, convert the volume to cm^3:
5.696 L = 5696 cm^3 (1L = 1000 cm^3)

Therefore, the volume of the spherical container is approximately 5696 cm^3.