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 When a portion of the helium is withdrawn and adjusted to 1.00 atm at 23°C, it is found to have a volume of 1.55 cm3.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
do i need to change the cc into ml?
for the equation?
am i using the 1 atm for pressure or the 1.950 atm for the first part?
for the first n i got .0638129 rounded... is that right?
The 638 part is right but the exponent is wrong. I suspect you used mL and not liters. Using PV = nRT,
P = 1 atm (that't the pressure of the 1.55 cc sample); V = 0.00155 Liters etc etc. I have 6.38 x 10^-5 mols for n. Check my work.
To calculate the volume of the spherical container, you can use the combined gas law equation:
(P1*V1)/T1 = (P2*V2)/T2
Where:
P1 = initial pressure of the gas in the container
V1 = initial volume of the gas in the container (unknown)
T1 = initial temperature of the gas in the container
P2 = pressure of the withdrawn portion of the gas
V2 = volume of the withdrawn portion of the gas
T2 = temperature of the withdrawn portion of the gas
Let's plug in the given values:
P1 = 1.950 atm
V1 = unknown
T1 = 25°C = 298K (converted to Kelvin)
P2 = 1.00 atm
V2 = 1.55 cm3 (converted to mL)
T2 = 23°C = 296K (converted to Kelvin)
Now, rearranging the equation to solve for V1:
V1 = (P2*V2*T1)/(P1*T2)
Let's substitute the values and calculate:
V1 = (1.00 atm * 1.55 cm3 * 298K) / (1.950 atm * 296K)
Converting cm3 to mL:
V1 = (1.00 atm * 1.55 mL * 298K) / (1.950 atm * 296K)
V1 = 1.55 * 298 / 1.950 = 238.97 mL
Therefore, the volume of the spherical container is approximately 238.97 mL.
Use PV = nRT to calculate n for the 1.55 cc removed.
Subtract P in the sphere of 1.950 - that after 1.55 cc were removed (1.71) and use PV =nRT to calculate volume.
Post your work if you get stuck.