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 When a portion of the helium is withdrawn and adjusted to 1.00 atm at 23°C, it is found to have a volume of 1.55 cm3.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
Did that value of 1.37 x 10^4 kg iron check for the problem we worked on last night.
yes it did, but now i can seem to figure out the 87% sulfuric...am i using the same number of moles as the Fe?
yes. Same number of mols. But you are using the wrong number of mols----WAY wrong. If you will post your work I can find where you are making an error. The number of mols is a huge number; that is, MUCH larger than what you showed in your calculations yesterday.
the spherical glass contains deeeeeek sample
To solve this problem, we can use the combined gas law formula. The combined gas law equation relates the initial and final conditions of a gas sample undergoing changes in temperature, pressure, and volume.
The combined gas law equation is given as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Here, P1 and P2 represent the initial and final pressures, V1 and V2 represent the initial and final volumes, and T1 and T2 represent the initial and final temperatures.
Let's assign the given values to the variables:
P1 = 1.950 atm
T1 = 25°C = 298 K
P2 = 1.00 atm
T2 = 23°C = 296 K
V2 = 1.55 cm^3
Now, we can rearrange the combined gas law equation to solve for V1, the volume of the spherical container:
V1 = (P2 * V2 * T1) / (P1 * T2)
Substituting the values, we have:
V1 = (1.00 atm * 1.55 cm^3 * 298 K) / (1.950 atm * 296 K)
Simplifying this expression, we get:
V1 ≈ 2.995 cm^3
Finally, to convert the volume to milliliters (mL), we know that 1 cm^3 is equal to 1 mL, so:
V1 ≈ 2.995 mL
Therefore, the volume of the spherical container is approximately 2.995 mL.