A sample of neon gas, Ne, has a density of g/L at a pressure of 1.13 atm and a temperature of 57°C?
The Correct Answer and Explanation is :
To determine the density of neon gas ((\text{Ne})) at a given pressure and temperature, we use the Ideal Gas Law in its density form:
[
\text{Density} (\rho) = \frac{PM}{RT}
]
Where:
- ( P ) = Pressure = 1.13 atm
- ( M ) = Molar mass of neon = 20.18 g/mol
- ( R ) = Ideal gas constant = 0.0821 L·atm/(mol·K)
- ( T ) = Temperature in Kelvin = 57°C + 273.15 = 330.15 K
Step 1: Plug in the values
[
\rho = \frac{(1.13 \text{ atm}) (20.18 \text{ g/mol})}{(0.0821 \text{ L·atm/(mol·K)}) (330.15 \text{ K})}
]
[
\rho = \frac{22.8034}{27.107415}
]
[
\rho \approx 0.841 \text{ g/L}
]
Explanation:
The Ideal Gas Law, ( PV = nRT ), describes the behavior of gases. To find density, we rearrange the equation in terms of mass per unit volume.
Since ( n = \frac{m}{M} ), replacing in ( PV = nRT ):
[
P V = \frac{m}{M} RT
]
Dividing both sides by ( V ):
[
P = \frac{m}{V} \frac{RT}{M}
]
Since density ( \rho = \frac{m}{V} ), the equation simplifies to:
[
\rho = \frac{PM}{RT}
]
Using this equation, we calculated that neon gas at 1.13 atm and 57°C has a density of 0.841 g/L.
This result aligns with the expected behavior of gases—higher temperature reduces density because the gas expands, while higher pressure increases density by compressing the gas.