Answer :
To calculate the density of argon gas ([tex]\(\text{Ar}(g)\)[/tex]) at [tex]\(-11^\circ \text{C}\)[/tex] and 675 mmHg, you can use the ideal gas law rearranged to find density. The ideal gas law is given by:
[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,
- [tex]\(T\)[/tex] is the temperature in Kelvin.
To find the density ([tex]\(d\)[/tex]), we need the mass of the gas per unit volume. The relation with the ideal gas law is:
[tex]\[ d = \frac{PM}{RT} \][/tex]
Where [tex]\(M\)[/tex] is the molar mass of the gas. Now, let's use the given values:
1. Convert Pressure:
- [tex]\( P = 675 \text{ mmHg} \)[/tex]
- Convert mmHg to atm: [tex]\( P = \frac{675}{760} \text{ atm} \approx 0.888 \text{ atm} \)[/tex]
2. Convert Temperature:
- [tex]\( T = -11^\circ C \)[/tex]
- Convert Celsius to Kelvin: [tex]\( T = -11 + 273.15 = 262.15 \text{ K} \)[/tex]
3. Molar Mass of Argon:
- [tex]\( M = 39.95 \text{ g/mol} \)[/tex]
4. Ideal Gas Constant (R):
- Use [tex]\( R = 62.3637 \text{ mmHg·L/mol·K} \)[/tex]
- To be consistent with our units (atm), convert this to [tex]\( R = 0.0821 \text{ atm·L/mol·K} \)[/tex]
5. Calculate Density:
- Substitute the values into the density formula:
[tex]\[
d = \frac{(0.888 \text{ atm}) \times (39.95 \text{ g/mol})}{(0.0821 \text{ atm·L/mol·K}) \times (262.15 \text{ K})}
\][/tex]
- By evaluating this, the density [tex]\(d\)[/tex] is approximately [tex]\(1.52 \text{ g/L}\)[/tex].
Therefore, the density of argon gas at [tex]\(-11^\circ \text{C}\)[/tex] and 675 mmHg is [tex]\(\textbf{1.52 g/L}\)[/tex]. The correct option is E) 1.52 g/L.
[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,
- [tex]\(T\)[/tex] is the temperature in Kelvin.
To find the density ([tex]\(d\)[/tex]), we need the mass of the gas per unit volume. The relation with the ideal gas law is:
[tex]\[ d = \frac{PM}{RT} \][/tex]
Where [tex]\(M\)[/tex] is the molar mass of the gas. Now, let's use the given values:
1. Convert Pressure:
- [tex]\( P = 675 \text{ mmHg} \)[/tex]
- Convert mmHg to atm: [tex]\( P = \frac{675}{760} \text{ atm} \approx 0.888 \text{ atm} \)[/tex]
2. Convert Temperature:
- [tex]\( T = -11^\circ C \)[/tex]
- Convert Celsius to Kelvin: [tex]\( T = -11 + 273.15 = 262.15 \text{ K} \)[/tex]
3. Molar Mass of Argon:
- [tex]\( M = 39.95 \text{ g/mol} \)[/tex]
4. Ideal Gas Constant (R):
- Use [tex]\( R = 62.3637 \text{ mmHg·L/mol·K} \)[/tex]
- To be consistent with our units (atm), convert this to [tex]\( R = 0.0821 \text{ atm·L/mol·K} \)[/tex]
5. Calculate Density:
- Substitute the values into the density formula:
[tex]\[
d = \frac{(0.888 \text{ atm}) \times (39.95 \text{ g/mol})}{(0.0821 \text{ atm·L/mol·K}) \times (262.15 \text{ K})}
\][/tex]
- By evaluating this, the density [tex]\(d\)[/tex] is approximately [tex]\(1.52 \text{ g/L}\)[/tex].
Therefore, the density of argon gas at [tex]\(-11^\circ \text{C}\)[/tex] and 675 mmHg is [tex]\(\textbf{1.52 g/L}\)[/tex]. The correct option is E) 1.52 g/L.