Answer :
To determine which container has the greatest pressure among the given four containers with respective volumes of 50 mL, 100 mL, 250 mL, and 500 mL, each containing 0.25 moles of helium gas at standard temperature (273.15 K), we can use the Ideal Gas Law. The Ideal Gas Law states:
[tex]\[ PV = nRT \][/tex]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant (62.364 L·mmHg/(mol·K) when dealing with volume in liters and pressure in mmHg),
- \( T \) is the temperature in Kelvin.
We need to find the pressure for each container. To start, let's convert the volumes from mL to L since \( R \) is given in terms of liters:
- 50 mL = 0.050 L
- 100 mL = 0.100 L
- 250 mL = 0.250 L
- 500 mL = 0.500 L
Now, applying the Ideal Gas Law to find the pressure in each container:
#### For 50 mL (0.050 L):
[tex]\[ P_1 = \frac{nRT}{V_1} = \frac{(0.25 \text{ moles}) \cdot (62.364 \text{ L·mmHg/(mol·K)}) \cdot (273.15 \text{ K})}{0.050 \text{ L}} \][/tex]
This yields:
[tex]\[ P_1 = 85173.633 \text{ mmHg} \][/tex]
#### For 100 mL (0.100 L):
[tex]\[ P_2 = \frac{nRT}{V_2} = \frac{(0.25 \text{ moles}) \cdot (62.364 \text{ L·mmHg/(mol·K)}) \cdot (273.15 \text{ K})}{0.100 \text{ L}} \][/tex]
This yields:
[tex]\[ P_2 = 42586.816 \text{ mmHg} \][/tex]
#### For 250 mL (0.250 L):
[tex]\[ P_3 = \frac{nRT}{V_3} = \frac{(0.25 \text{ moles}) \cdot (62.364 \text{ L·mmHg/(mol·K)}) \cdot (273.15 \text{ K})}{0.250 \text{ L}} \][/tex]
This yields:
[tex]\[ P_3 = 17034.727 \text{ mmHg} \][/tex]
#### For 500 mL (0.500 L):
[tex]\[ P_4 = \frac{nRT}{V_4} = \frac{(0.25 \text{ moles}) \cdot (62.364 \text{ L·mmHg/(mol·K)}) \cdot (273.15 \text{ K})}{0.500 \text{ L}} \][/tex]
This yields:
[tex]\[ P_4 = 8517.363 \text{ mmHg} \][/tex]
By inspecting the calculated pressures, we can see:
- \( P_1 = 85173.633 \) mmHg
- \( P_2 = 42586.816 \) mmHg
- \( P_3 = 17034.727 \) mmHg
- \( P_4 = 8517.363 \) mmHg
The container with the 50 mL volume has the greatest pressure of 85173.633 mmHg, which is the highest among the calculated pressures.
Thus, the container that would have the greatest pressure is:
A. 50 mL
[tex]\[ PV = nRT \][/tex]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant (62.364 L·mmHg/(mol·K) when dealing with volume in liters and pressure in mmHg),
- \( T \) is the temperature in Kelvin.
We need to find the pressure for each container. To start, let's convert the volumes from mL to L since \( R \) is given in terms of liters:
- 50 mL = 0.050 L
- 100 mL = 0.100 L
- 250 mL = 0.250 L
- 500 mL = 0.500 L
Now, applying the Ideal Gas Law to find the pressure in each container:
#### For 50 mL (0.050 L):
[tex]\[ P_1 = \frac{nRT}{V_1} = \frac{(0.25 \text{ moles}) \cdot (62.364 \text{ L·mmHg/(mol·K)}) \cdot (273.15 \text{ K})}{0.050 \text{ L}} \][/tex]
This yields:
[tex]\[ P_1 = 85173.633 \text{ mmHg} \][/tex]
#### For 100 mL (0.100 L):
[tex]\[ P_2 = \frac{nRT}{V_2} = \frac{(0.25 \text{ moles}) \cdot (62.364 \text{ L·mmHg/(mol·K)}) \cdot (273.15 \text{ K})}{0.100 \text{ L}} \][/tex]
This yields:
[tex]\[ P_2 = 42586.816 \text{ mmHg} \][/tex]
#### For 250 mL (0.250 L):
[tex]\[ P_3 = \frac{nRT}{V_3} = \frac{(0.25 \text{ moles}) \cdot (62.364 \text{ L·mmHg/(mol·K)}) \cdot (273.15 \text{ K})}{0.250 \text{ L}} \][/tex]
This yields:
[tex]\[ P_3 = 17034.727 \text{ mmHg} \][/tex]
#### For 500 mL (0.500 L):
[tex]\[ P_4 = \frac{nRT}{V_4} = \frac{(0.25 \text{ moles}) \cdot (62.364 \text{ L·mmHg/(mol·K)}) \cdot (273.15 \text{ K})}{0.500 \text{ L}} \][/tex]
This yields:
[tex]\[ P_4 = 8517.363 \text{ mmHg} \][/tex]
By inspecting the calculated pressures, we can see:
- \( P_1 = 85173.633 \) mmHg
- \( P_2 = 42586.816 \) mmHg
- \( P_3 = 17034.727 \) mmHg
- \( P_4 = 8517.363 \) mmHg
The container with the 50 mL volume has the greatest pressure of 85173.633 mmHg, which is the highest among the calculated pressures.
Thus, the container that would have the greatest pressure is:
A. 50 mL
