Calculate the volume, in liters, of 5.80×10−3 mol of a gas at 44 ∘C and 0.885 atm .
p v = n r t
convert ºC to ºK and make sure your r value has the correct units
... liters, moles, ºK, atm
1 mole occupies 22.4L at STP
You can use PV=nRT or just realize that
PV = kT, so
PV/T = k
For 1 mole, you want V such that
(1*22.4)/(273.15) = (0.885*V)/(273.15+44)
V = 29.3879 L
But you only have 0.0058 moles, so multiply V by that, and you get
V = 0.170L or 170mL
To calculate the volume of a gas, we can use the ideal gas law formula:
PV = nRT
where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (K)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 44 + 273.15 = 317.15 K
Now we can substitute the values into the ideal gas law equation and solve for V:
PV = nRT
V = (nRT) / P
V = (5.80×10^(-3) mol) * (0.0821 L·atm/(mol·K)) * (317.15 K) / (0.885 atm)
Calculating this expression will give us the volume in liters.