Exactly 500 cm3 of nitrogen is collected over water at 25°C and 755 mmHg. The gas is saturated with water
vapour. Calculate the volume of the nitrogen in the dry condition at 0 °C and 1 atm. Vapour pressure of water at
25°C is 23.8 mmHg.
My working is:
P =1.0247 atm (converting mmHg to atm), V=0.5L, T=323K (25+25+273), R=0.08206.
Using PV=nRT, I solved for n, which is 0.01933. Using this formula again, with the other info, 1atm(V)=0.01933(0.08206)(273K) and I get V=433 cm^3. The answer says 441cm^3 and I don't know what I could have possibly done wrong. Any help is much appreciated.
n=(755-23.8)*.500/(8.31*298)= 0.147634854 moles nitrogen gas
Now, at STP, now water vapor, that translates tso
V= nRT/P= 0.147634854*8.13*273/(755-23.8)=.448 dm^3 so I agree with you.
Your calculation seems almost correct. Let's review the steps and see where the discrepancy may have occurred.
First, let's convert the given pressure values to atm:
755 mmHg = 755/760 atm ≈ 0.9947 atm
23.8 mmHg = 23.8/760 atm ≈ 0.0313 atm
Next, we can calculate the partial pressure of nitrogen gas:
P_nitrogen = Total pressure - Vapour pressure of water
P_nitrogen = 0.9947 atm - 0.0313 atm ≈ 0.9634 atm
Now, let's use the ideal gas law equation, PV = nRT, to find the number of moles of nitrogen:
P_nitrogen * V_nitrogen = n * R * T
Substituting the known values:
0.9634 atm * V_nitrogen = 0.01933 mol * 0.08206 L·atm/(mol·K) * 298 K
Simplifying:
V_nitrogen = (0.01933 mol * 0.08206 L·atm/(mol·K) * 298 K) / 0.9634 atm
V_nitrogen ≈ 0.442 L ≈ 442 cm^3
So, the volume of nitrogen gas in the dry condition at 0°C and 1 atm should be approximately 442 cm^3, not 441 cm^3 as stated in the answer. It seems like a rounding error in the answer key. Therefore, your calculation was correct, and there doesn't appear to be an error in your working.
To solve this problem, we need to use the ideal gas law equation, PV = nRT, where:
- P = pressure of the gas
- V = volume of the gas
- n = number of moles of the gas
- R = ideal gas constant
- T = temperature of the gas
First, let's calculate the number of moles of nitrogen gas using the given conditions:
P1 = 755 mmHg = 0.993 atm
V1 = 500 cm^3 = 0.5 L
T1 = 25°C = 298 K
R = 0.08206 L·atm/(mol·K)
We can rearrange the ideal gas law equation to solve for n:
n = (P1 * V1) / (R * T1)
= (0.993 atm * 0.5 L) / (0.08206 L·atm/(mol·K) * 298 K)
≈ 0.01929 moles
Now, let's convert this number of moles to the volume at the new conditions:
P2 = 1 atm
V2 = ?
T2 = 0°C = 273 K
We can rearrange the ideal gas law equation again, this time solving for V2:
V2 = (n * R * T2) / P2
= (0.01929 moles * 0.08206 L·atm/(mol·K) * 273 K) / 1 atm
≈ 0.4317 L
Finally, let's convert the volume from liters to cm^3:
V2_cm3 = 0.4317 L * 1000 cm^3/L
≈ 431.7 cm^3
So, the volume of nitrogen gas in dry condition at 0°C and 1 atm pressure is approximately 431.7 cm^3.
If the correct answer is stated as 441 cm^3, there might be a small rounding error or difference in the value of the gas constant used. Double-check the values you used and the significant figures in your calculations to see if any minor discrepancies might account for the difference in the final result.