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 When a portion of the helium is withdrawn and adjusted to \(1.00\) atm at \(25^{\circ} \mathrm{C}\), it is found to have a volume of \(1.75 \mathrm{~cm}^{3}\).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

Temperature in Celsius should be converted to Kelvin by adding 273.15. Therefore, Temperature in Kelvin, T = \(\displaystyle 25^{\circ}\mathrm{C}\) + 273.15 = 298.15 K

We will use the Ideal Gas Law equation to find the number of moles for the withdrawn helium and the remaining helium: a) For the withdrawn helium (with given volume): P = 1.00 atm V = 1.75 cm³ = 0.00175 L (Converting to Liters) T = 298.15 K R = 0.0821 L atm K⁻¹ mol⁻¹ (Ideal Gas Constant) To find the number of moles 'n', we use the equation: PV = nRT \(n_\textrm{withdrawn} = \dfrac{PV}{RT} = \dfrac{(1.00\textrm{ atm})(0.00175\textrm{ L})}{(0.0821\textrm{ L atm K}^{-1}\textrm{ mol}^{-1})(298.15\textrm{ K})}\) Now, let's calculate the number of moles withdrawn: \(n_\textrm{withdrawn} = 0.0000727\textrm{ mol}\) b) For the remaining helium (assuming withdrawn and remaining helium added to the initial total pressure): P = 1.710 atm (new pressure after withdrawing helium) V remains unknown T = 298.15 K R = 0.0821 L atm K⁻¹ mol⁻¹ (Ideal Gas Constant) To find the number of moles 'n', we use the equation: PV = nRT \(n_\textrm{remaining} = \dfrac{PV}{RT} = \dfrac{(1.710\textrm{ atm})(V)}{(0.0821\textrm{ L atm K}^{-1}\textrm{ mol}^{-1})(298.15\textrm{ K})}\) We will leave this equation as-is for now since we don't know the volume V.

We now add the number of moles of the withdrawn helium and the remaining helium in the container: \(n_\textrm{total} = n_\textrm{withdrawn}+n_\textrm{remaining}\) As we don't know the number of moles of the remaining helium, we will use the expression found above for the remaining helium: \(n_\textrm{total} = 0.0000727\textrm{ mol}+\dfrac{(1.710\textrm{ atm})(V)}{(0.0821\textrm{ L atm K}^{-1}\textrm{ mol}^{-1})(298.15\textrm{ K})}\)

Now we write the Ideal Gas Law for the total helium before the amount was withdrawn (Applying initial pressure given 1.960 atm): \(PV = nRT\) \((1.960\textrm{ atm})(V) = n_\textrm{total}(0.0821\textrm{ L atm K}^{-1}\textrm{ mol}^{-1})(298.15\textrm{ K})\) Now, we substitute the expression found for \(n_\textrm{total}\) (Step 3) into this equation: \((1.960\textrm{ atm})(V) = \left[ 0.0000727\textrm{ mol}+\dfrac{(1.710\textrm{ atm})(V)}{(0.0821\textrm{ L atm K}^{-1}\textrm{ mol}^{-1})(298.15\textrm{ K})}\right](0.0821\textrm{ L atm K}^{-1}\textrm{ mol}^{-1})(298.15\textrm{ K})\) Now we will solve this equation for V, the volume of the spherical glass container: \(V = 0.00460 \textrm{ L}\) To convert this to cm³, we multiply by 1000: Volume of the spherical glass container = 4.60 cm³