I'm trying to find the dry gas pressure of H2 in mmHg. The barometric pressure is 765.3 mmHg, the vp of H20 is 23.76 mmHg and the gas collected was 25.6 mL.
this is what i did.
765.3 mmHg - 23.76 mmHg - (25.6 Ml/13.6 mmHg) = 739.66 = 740 mmHg 93 sig figs).
Then I have to use the combined gas law to convert this volume of the H2 gas collected to the new volume at STP.
So i used the formula v2 = P1 V1 T2/ T1 P1
I converted the volume of H2 into atm and got .974 atm (740 mmHg/760). Then I multiplied .974 atm by .0244 L (the leftover liquid) and 298 K (the temp.) then I divided by 273 K and 1 atm.
(.974 atm x .0244 L x 298 K)/(273 K x 1 atm) = .0295 L
Is that correct?
The dry gas pressure is pressure inside the tube - watervapor pressure. The pressure has nothing to do with the volume collected, unless you are supporting a column of water, then you have to adjust barometric pressure for the height of the column, it does not appear you had to do this correction.
The rest of it is correct, if dry pressure is corrected.
Let's break down the calculation step by step to determine if your approach is correct.
First, you correctly subtracted the vapor pressure of water (23.76 mmHg) from the barometric pressure (765.3 mmHg) to find the partial pressure of dry gas (H2). So, the pressure of the dry gas is 765.3 mmHg - 23.76 mmHg = 741.54 mmHg. You rounded it to 740 mmHg, which has three significant figures.
Next, you mentioned using the combined gas law to convert the volume of the H2 gas collected to the new volume at STP (Standard Temperature and Pressure). However, the formula you used seems to be incorrect. The correct formula for the combined gas law is:
(P1 * V1 * T2) / (T1 * P2) = V2
Here, P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2 and T2 are the final pressure and temperature.
Let's assume that the initial temperature (T1) is equal to the final temperature (T2), and both are 298 K (25 degrees Celsius). Additionally, since we are trying to calculate the new volume at STP, the final pressure (P2) should be 1 atm (760 mmHg). Given these values, we can rewrite the combined gas law equation as:
(P1 * V1) / P2 = V2
Now, let's substitute the values you have:
(P1 * V1) / P2 = V2
(740 mmHg * 25.6 mL) / 760 mmHg = V2
Before calculating, remember to convert the volume from mL to L, as the pressure is in mmHg. So, 25.6 mL is equal to 0.0256 L.
(740 mmHg * 0.0256 L) / 760 mmHg = V2
19.04 L / 760 mmHg = V2
Now, to convert mmHg to atm, divide the pressure by 760 mmHg/1 atm:
19.04 L / (760 mmHg / 1 atm) = V2
19.04 L / 760 mmHg * 1 atm = V2
0.025 L = V2
So, the final volume of the dry H2 gas at STP is 0.025 L or 25 mL (since 1 L = 1000 mL).
To summarize:
1. The dry gas pressure of H2 is approximately 740 mmHg.
2. The final volume of H2 gas collected at STP is 25 mL.
Please note that the calculations are based on the assumptions and values you provided. Make sure to double-check the values and units to ensure accuracy.