Determine the number of moles of air present in 1.35 L at 100. kPa and 23.0°C (R = 8.317 liter•kPa/mol•K
a) 0.0548 mol
b) 205 mol
c) the number of moles is held constant
d) 0.706 mol
Please help! I can't get this!!
Use PV = nRT. Substitute and solve for n = number of mols air. Remember to use K for temperature. T in Kelvin = 23.0 C + 273.15 = ?
Show your work if you are stuck.
To determine the number of moles of air present in 1.35 L at 100 kPa and 23.0°C, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure (in kPa)
V = volume (in liters)
n = number of moles
R = gas constant (8.317 liter•kPa/mol•K)
T = temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin. The conversion equation is:
T(K) = T(°C) + 273.15
Substituting the given values:
T(K) = 23.0 + 273.15
T(K) = 296.15 K
Now, we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the given values:
n = (100 kPa) * (1.35 L) / (8.317 liter•kPa/mol•K * 296.15 K)
Simplifying:
n ≈ 0.0548 mol
Therefore, the number of moles of air present in 1.35 L at 100 kPa and 23.0°C is approximately 0.0548 mol. Hence, the correct option is a) 0.0548 mol.
To determine the number of moles of air present in the given volume and conditions, we can use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
T (K) = 23°C + 273.15 = 296.15 K
Next, we rearrange the Ideal Gas Law equation to solve for the number of moles (n):
n = PV / RT
Now, plug in the given values:
P = 100. kPa
V = 1.35 L
R = 8.317 liter•kPa/mol•K
T = 296.15 K
Substituting these values into the equation, we have:
n = (100. kPa)(1.35 L) / (8.317 liter•kPa/mol•K)(296.15 K)
Now, let's calculate the number of moles:
n ≈ 0.05478 mol
Approximating to the correct number of significant figures, the answer is approximately 0.0548 mol. Therefore, the correct option is a) 0.0548 mol.