Answer :
To find the temperature of the canister containing 75.0 liters of argon gas with 15.82 moles at a pressure of 546.8 kilopascals, we can use the ideal gas law, which is given by the equation:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure in pascals (Pa)
- [tex]\( V \)[/tex] is the volume in cubic meters (m³)
- [tex]\( n \)[/tex] is the number of moles
- [tex]\( R \)[/tex] is the ideal gas constant
- [tex]\( T \)[/tex] is the temperature in Kelvin
First, let's convert the given pressure from kilopascals to pascals:
[tex]\[ 546.8 \text{ kPa} = 546.8 \times 1000 \text{ Pa} = 546800 \text{ Pa} \][/tex]
Next, note that the volume is already in liters, so we convert it to cubic meters (since 1 liter = 0.001 cubic meters):
[tex]\[ 75.0 \text{ liters} = 75.0 \times 0.001 \text{ m}^3 = 0.075 \text{ m}^3 \][/tex]
We'll use the universal gas constant [tex]\( R \)[/tex] in the appropriate units (J/(mol K)):
[tex]\[ R = 8.314 \text{ J/(mol K)} \][/tex]
Now, we can rearrange the ideal gas law to solve for the temperature [tex]\( T \)[/tex]:
[tex]\[ T = \frac{PV}{nR} \][/tex]
Substitute known values into the equation:
[tex]\[ T = \frac{(546800 \text{ Pa}) \times (0.075 \text{ m}^3)}{(15.82 \text{ mol}) \times (8.314 \text{ J/(mol K)})} \][/tex]
[tex]\[ T = \frac{41010 \text{ (Pa m}^3)}{131.50948 \text{ (mol J/(mol K))}} \][/tex]
[tex]\[ T = 311797.9603958047 \text{ K} \][/tex]
Rounding this to three significant figures:
[tex]\[ T \approx 3.12 \times 10^5 \text{ K} \][/tex]
Thus, the temperature of the canister is [tex]\( 3.12 \times 10^5 \)[/tex] K.
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure in pascals (Pa)
- [tex]\( V \)[/tex] is the volume in cubic meters (m³)
- [tex]\( n \)[/tex] is the number of moles
- [tex]\( R \)[/tex] is the ideal gas constant
- [tex]\( T \)[/tex] is the temperature in Kelvin
First, let's convert the given pressure from kilopascals to pascals:
[tex]\[ 546.8 \text{ kPa} = 546.8 \times 1000 \text{ Pa} = 546800 \text{ Pa} \][/tex]
Next, note that the volume is already in liters, so we convert it to cubic meters (since 1 liter = 0.001 cubic meters):
[tex]\[ 75.0 \text{ liters} = 75.0 \times 0.001 \text{ m}^3 = 0.075 \text{ m}^3 \][/tex]
We'll use the universal gas constant [tex]\( R \)[/tex] in the appropriate units (J/(mol K)):
[tex]\[ R = 8.314 \text{ J/(mol K)} \][/tex]
Now, we can rearrange the ideal gas law to solve for the temperature [tex]\( T \)[/tex]:
[tex]\[ T = \frac{PV}{nR} \][/tex]
Substitute known values into the equation:
[tex]\[ T = \frac{(546800 \text{ Pa}) \times (0.075 \text{ m}^3)}{(15.82 \text{ mol}) \times (8.314 \text{ J/(mol K)})} \][/tex]
[tex]\[ T = \frac{41010 \text{ (Pa m}^3)}{131.50948 \text{ (mol J/(mol K))}} \][/tex]
[tex]\[ T = 311797.9603958047 \text{ K} \][/tex]
Rounding this to three significant figures:
[tex]\[ T \approx 3.12 \times 10^5 \text{ K} \][/tex]
Thus, the temperature of the canister is [tex]\( 3.12 \times 10^5 \)[/tex] K.
