Answer :
To find the density of Argon gas ([tex]\(\operatorname{Ar}(g)\)[/tex]) at [tex]\(-11^\circ \text{C}\)[/tex] and a pressure of 675 mmHg, we can use the Ideal Gas Law in the form suitable for density calculation.
Step-by-step Solution:
1. Understand the Ideal Gas Law:
The Ideal Gas Law is given by the equation:
[tex]\[
PV = nRT
\][/tex]
where [tex]\( P \)[/tex] is the pressure, [tex]\( V \)[/tex] is the volume, [tex]\( n \)[/tex] is the number of moles, [tex]\( R \)[/tex] is the ideal gas constant, and [tex]\( T \)[/tex] is the temperature in Kelvin.
2. Make the necessary substitutions for density:
Density ([tex]\(\rho\)[/tex]) is mass per unit volume:
[tex]\[
\rho = \frac{m}{V}
\][/tex]
Using the molar mass [tex]\( M \)[/tex] and solving for density, the relationship becomes:
[tex]\[
\rho = \frac{PM}{RT}
\][/tex]
where [tex]\( M \)[/tex] is the molar mass of Argon.
3. Convert temperature to Kelvin:
Temperature in Kelvin ([tex]\( T \)[/tex]) is needed for the gas laws. Convert from Celsius:
[tex]\[
T = -11 + 273.15 = 262.15 \text{ K}
\][/tex]
4. Use appropriate values:
- Pressure [tex]\( P = 675 \text{ mmHg} \)[/tex]
- Molar mass of Argon [tex]\( M = 39.948 \text{ g/mol} \)[/tex]
- Ideal gas constant [tex]\( R = 62.36 \text{ L mmHg/mol K} \)[/tex]
5. Calculate the density:
Substitute these values into the density formula:
[tex]\[
\rho = \frac{675 \times 39.948}{62.36 \times 262.15}
\][/tex]
After performing the calculation, you find:
[tex]\[
\rho \approx 1.65 \text{ g/L}
\][/tex]
6. Conclusion:
The density of Argon gas at [tex]\(-11^\circ \text{C}\)[/tex] and 675 mmHg is approximately [tex]\(1.65 \text{ g/L}\)[/tex]. Therefore, the correct choice from the given options is [tex]\(1.65 \text{ g/L}\)[/tex].
Step-by-step Solution:
1. Understand the Ideal Gas Law:
The Ideal Gas Law is given by the equation:
[tex]\[
PV = nRT
\][/tex]
where [tex]\( P \)[/tex] is the pressure, [tex]\( V \)[/tex] is the volume, [tex]\( n \)[/tex] is the number of moles, [tex]\( R \)[/tex] is the ideal gas constant, and [tex]\( T \)[/tex] is the temperature in Kelvin.
2. Make the necessary substitutions for density:
Density ([tex]\(\rho\)[/tex]) is mass per unit volume:
[tex]\[
\rho = \frac{m}{V}
\][/tex]
Using the molar mass [tex]\( M \)[/tex] and solving for density, the relationship becomes:
[tex]\[
\rho = \frac{PM}{RT}
\][/tex]
where [tex]\( M \)[/tex] is the molar mass of Argon.
3. Convert temperature to Kelvin:
Temperature in Kelvin ([tex]\( T \)[/tex]) is needed for the gas laws. Convert from Celsius:
[tex]\[
T = -11 + 273.15 = 262.15 \text{ K}
\][/tex]
4. Use appropriate values:
- Pressure [tex]\( P = 675 \text{ mmHg} \)[/tex]
- Molar mass of Argon [tex]\( M = 39.948 \text{ g/mol} \)[/tex]
- Ideal gas constant [tex]\( R = 62.36 \text{ L mmHg/mol K} \)[/tex]
5. Calculate the density:
Substitute these values into the density formula:
[tex]\[
\rho = \frac{675 \times 39.948}{62.36 \times 262.15}
\][/tex]
After performing the calculation, you find:
[tex]\[
\rho \approx 1.65 \text{ g/L}
\][/tex]
6. Conclusion:
The density of Argon gas at [tex]\(-11^\circ \text{C}\)[/tex] and 675 mmHg is approximately [tex]\(1.65 \text{ g/L}\)[/tex]. Therefore, the correct choice from the given options is [tex]\(1.65 \text{ g/L}\)[/tex].
