Answer:So, the mass of the gas sample under room conditions is approximately 0.2802 grams.
Step-by-step explanation:
To find the volume and mass of the sample of gas under room conditions (typically around 298 K and 1 atm), you can use the ideal gas law, which is expressed as:
PV = nRT
Where:
P = Pressure (in atmospheres, atm)
V = Volume (in liters, L)
n = Number of moles of gas (in moles, mol)
R = Universal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin, K)
First, you need to convert the given temperature from Celsius to Kelvin using the formula:
T(K) = T(°C) + 273.15
So, for a temperature of 2.0°C:
T(K) = 2.0°C + 273.15 = 275.15 K
Now, you can rearrange the ideal gas law to solve for n (moles of gas):
n = (PV) / (RT)
You have:
P = 2.4 atm
V = 1.2 x 10^-3 dm³ = 1.2 x 10^-6 L (since 1 dm³ = 1 L)
T = 275.15 K
R = 0.0821 L·atm/(mol·K)
Now, plug these values into the equation to calculate n:
n = (2.4 atm * 1.2 x 10^-6 L) / (0.0821 L·atm/(mol·K) * 275.15 K)
Calculate n:
n ≈ 1.001 x 10^-5 mol
Now that you have the number of moles (n) of the gas, you can calculate the volume (V) at room conditions (298 K and 1 atm) using the same ideal gas law:
V = (nRT) / P
Plug in the values:
V = (1.001 x 10^-5 mol * 0.0821 L·atm/(mol·K) * 298 K) / 1 atm
Calculate V:
V ≈ 0.247 L
So, the volume of the gas sample under room conditions is approximately 0.247 liters.
To calculate the mass (m) of the gas sample, you can use the formula:
m = n * M
Where:
m = mass (in grams, g)
n = number of moles (from earlier calculations)
M = molar mass of the gas (in g/mol)
Since you haven't specified the gas, let's assume it's a common gas like nitrogen (N2), which has a molar mass of approximately 28.02 g/mol.
Calculate the mass:
m = (1.001 x 10^-5 mol) * (28.02 g/mol) ≈ 0.2802 g