Answer :
To find the density of chlorine gas under the given conditions, we can use the Ideal Gas Law in the modified form that relates density to pressure, molar mass, the gas constant, and temperature:
The Ideal Gas Law is [tex]\( PV = nRT \)[/tex].
First, let's convert the given temperature from Celsius to Kelvin:
[tex]\[ T(K) = 25.0°C + 273.15 = 298.15\,K \][/tex]
We need the universal gas constant in the appropriate units (L·atm·K⁻¹·mol⁻¹):
[tex]\[ R = 0.0821\, \text{L·atm·K⁻¹·mol⁻¹} \][/tex]
The density ([tex]\( \rho \)[/tex]) formula derived from the Ideal Gas Law is:
[tex]\[ \rho = \frac{PM}{RT} \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure (1.50 atm),
- [tex]\( M \)[/tex] is the molar mass (71.0 g/mol),
- [tex]\( R \)[/tex] is the universal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹),
- [tex]\( T \)[/tex] is the temperature in Kelvin (298.15 K).
Substitute the known values into the formula:
[tex]\[ \rho = \frac{(1.50\, \text{atm})(71.0\, \text{g/mol})}{(0.0821\, \text{L·atm·K⁻¹·mol⁻¹})(298.15\, \text{K})} \][/tex]
Perform the calculation step by step:
1. Calculate the numerator:
[tex]\[ 1.50\, \text{atm} \times 71.0\, \text{g/mol} = 106.5\, \text{g·atm/mol} \][/tex]
2. Calculate the denominator:
[tex]\[ 0.0821\, \text{L·atm·K⁻¹·mol⁻¹} \times 298.15\, \text{K} = 24.487515\, \text{L·atm/mol} \][/tex]
3. Divide the numerator by the denominator:
[tex]\[ \rho = \frac{106.5}{24.487515} \approx 4.349 \][/tex]
Finally, round the answer to three significant figures:
[tex]\[ \rho \approx 4.35 \][/tex]
The density of chlorine gas is approximately 4.35 g/L.
The Ideal Gas Law is [tex]\( PV = nRT \)[/tex].
First, let's convert the given temperature from Celsius to Kelvin:
[tex]\[ T(K) = 25.0°C + 273.15 = 298.15\,K \][/tex]
We need the universal gas constant in the appropriate units (L·atm·K⁻¹·mol⁻¹):
[tex]\[ R = 0.0821\, \text{L·atm·K⁻¹·mol⁻¹} \][/tex]
The density ([tex]\( \rho \)[/tex]) formula derived from the Ideal Gas Law is:
[tex]\[ \rho = \frac{PM}{RT} \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure (1.50 atm),
- [tex]\( M \)[/tex] is the molar mass (71.0 g/mol),
- [tex]\( R \)[/tex] is the universal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹),
- [tex]\( T \)[/tex] is the temperature in Kelvin (298.15 K).
Substitute the known values into the formula:
[tex]\[ \rho = \frac{(1.50\, \text{atm})(71.0\, \text{g/mol})}{(0.0821\, \text{L·atm·K⁻¹·mol⁻¹})(298.15\, \text{K})} \][/tex]
Perform the calculation step by step:
1. Calculate the numerator:
[tex]\[ 1.50\, \text{atm} \times 71.0\, \text{g/mol} = 106.5\, \text{g·atm/mol} \][/tex]
2. Calculate the denominator:
[tex]\[ 0.0821\, \text{L·atm·K⁻¹·mol⁻¹} \times 298.15\, \text{K} = 24.487515\, \text{L·atm/mol} \][/tex]
3. Divide the numerator by the denominator:
[tex]\[ \rho = \frac{106.5}{24.487515} \approx 4.349 \][/tex]
Finally, round the answer to three significant figures:
[tex]\[ \rho \approx 4.35 \][/tex]
The density of chlorine gas is approximately 4.35 g/L.
