Question
Hydrogen sulfide, H2S, has the characteristic odor of rotten eggs. If a sample of H2S gas at 743.2 mmg and 204.8 K in a 30.0 L container is allowed to enter a 89.5 L container at 496.0 K, what is the pressure in the new container? Be sure your answer has the correct number of significant figures.
Answers
GPT-4o mini
To find the pressure of hydrogen sulfide (H₂S) in the new container, we can use the ideal gas law and the principles of gas behavior during the transfer.
### Step 1: Calculate the number of moles of H₂S in the initial container.
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles (in moles)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)
First, we need to convert the initial pressure from mmHg to atm:
\[
P = \frac{743.2 \text{ mmHg}}{760 \text{ mmHg/atm}} \approx 0.976 atm
\]
Now we can calculate the number of moles \( n \) in the initial container:
\[
n = \frac{PV}{RT} = \frac{(0.976 \text{ atm}) \cdot (30.0 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)}) \cdot (204.8 \text{ K})}
\]
Calculating \( n \):
\[
n = \frac{(0.976)(30.0)}{(0.0821)(204.8)} \approx \frac{29.28}{16.82568} \approx 1.743 mol
\]
### Step 2: Calculate the pressure in the new container.
The new container volume is 89.5 L, and the temperature is 496.0 K. We'll use the ideal gas law again to find the new pressure \( P' \):
\[
P' = \frac{nRT'}{V'}
\]
Where:
- \( T' = 496.0 \text{ K} \)
- \( V' = 89.5 \text{ L} \)
Substituting in the values we have calculated for \( n \) and using the ideal gas constant:
\[
P' = \frac{(1.743 \text{ mol}) \cdot (0.0821 \text{ L·atm/(K·mol)}) \cdot (496.0 \text{ K})}{89.5 \text{ L}}
\]
Calculating \( P' \):
\[
P' = \frac{(1.743)(0.0821)(496.0)}{89.5} \approx \frac{(1.743)(40.7926)}{89.5} \approx \frac{71.03}{89.5} \approx 0.793 atm
\]
### Step 3: Convert to mmHg for clarity and use the correct significant figures.
To get the pressure in mmHg:
\[
P' \approx 0.793 \text{ atm} \cdot 760 \text{ mmHg/atm} \approx 603.68 \text{ mmHg}
\]
### Final Answer:
Considering significant figures, we have three significant figures from the initial conditions. Therefore, the final pressure in the new container, rounded to three significant figures, is:
\[\boxed{604 \text{ mmHg}}\]
### Step 1: Calculate the number of moles of H₂S in the initial container.
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles (in moles)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)
First, we need to convert the initial pressure from mmHg to atm:
\[
P = \frac{743.2 \text{ mmHg}}{760 \text{ mmHg/atm}} \approx 0.976 atm
\]
Now we can calculate the number of moles \( n \) in the initial container:
\[
n = \frac{PV}{RT} = \frac{(0.976 \text{ atm}) \cdot (30.0 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)}) \cdot (204.8 \text{ K})}
\]
Calculating \( n \):
\[
n = \frac{(0.976)(30.0)}{(0.0821)(204.8)} \approx \frac{29.28}{16.82568} \approx 1.743 mol
\]
### Step 2: Calculate the pressure in the new container.
The new container volume is 89.5 L, and the temperature is 496.0 K. We'll use the ideal gas law again to find the new pressure \( P' \):
\[
P' = \frac{nRT'}{V'}
\]
Where:
- \( T' = 496.0 \text{ K} \)
- \( V' = 89.5 \text{ L} \)
Substituting in the values we have calculated for \( n \) and using the ideal gas constant:
\[
P' = \frac{(1.743 \text{ mol}) \cdot (0.0821 \text{ L·atm/(K·mol)}) \cdot (496.0 \text{ K})}{89.5 \text{ L}}
\]
Calculating \( P' \):
\[
P' = \frac{(1.743)(0.0821)(496.0)}{89.5} \approx \frac{(1.743)(40.7926)}{89.5} \approx \frac{71.03}{89.5} \approx 0.793 atm
\]
### Step 3: Convert to mmHg for clarity and use the correct significant figures.
To get the pressure in mmHg:
\[
P' \approx 0.793 \text{ atm} \cdot 760 \text{ mmHg/atm} \approx 603.68 \text{ mmHg}
\]
### Final Answer:
Considering significant figures, we have three significant figures from the initial conditions. Therefore, the final pressure in the new container, rounded to three significant figures, is:
\[\boxed{604 \text{ mmHg}}\]