Let's solve this step-by-step using the ideal gas law, [tex]\( PV = nRT \)[/tex].
1. Identify the given values:
- Volume, [tex]\( V = 75.0 \)[/tex] liters
- Moles of argon, [tex]\( n = 15.82 \)[/tex] moles
- Pressure, [tex]\( P = 546.8 \)[/tex] kPa
- Ideal gas constant, [tex]\( R = 8.314 \frac{L\cdot kPa}{mol\cdot K} \)[/tex]
2. Rearrange the ideal gas law to solve for temperature [tex]\( T \)[/tex]:
[tex]\[
PV = nRT \implies T = \frac{PV}{nR}
\][/tex]
3. Substitute the given values into the equation:
[tex]\[
T = \frac{(546.8 \, \text{kPa}) \times (75.0 \, \text{L})}{(15.82 \, \text{moles}) \times (8.314 \frac{L\cdot kPa}{mol\cdot K})}
\][/tex]
4. Perform the calculation.
[tex]\[
T \approx 311.798 \, \text{K}
\][/tex]
5. Express the answer to three significant figures:
[tex]\[
T \approx 311.798 \rightarrow 311.798 \, \text{K}
\][/tex]
Therefore, the temperature of the canister is [tex]\( 311.798 \)[/tex] K.
