What volume (in liters) is occupied by 0.01 moles of carbon monoxide, CO at 9 degrees C and 0.973 atm?
PV = nRT
pv=nrt
hence, v=nrt/p
v=(0.01*0.082*282)10.972
v= 0.238L
To find the volume occupied by 0.01 moles of carbon monoxide (CO) at a given temperature and pressure, you can use the ideal gas law formula:
PV = nRT
Where:
P = pressure in atmospheres (atm)
V = volume in liters (L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (K)
First, we need to convert the temperature from Celsius to Kelvin.
9 degrees Celsius + 273.15 = 282.15 K
Now, we can rearrange the formula to solve for V:
V = (nRT) / P
Substituting the values we have:
V = (0.01 moles x 0.0821 L·atm/(mol·K) x 282.15 K) / 0.973 atm
V = (0.00821 L·atm x 282.15 K) / 0.973 atm
V ≈ 2.38 L
Therefore, approximately 2.38 liters of carbon monoxide (CO) would be occupied by 0.01 moles of CO at 9 degrees C and 0.973 atm.
To find the volume (in liters) occupied by a given amount of gas, we can use the ideal gas law equation:
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)
First, we need to convert the temperature from degrees Celsius to Kelvin:
T(K) = T(°C) + 273.15
So, the temperature in Kelvin will be:
T = 9 + 273.15 = 282.15 K
Now, we can rearrange the ideal gas law equation to solve for volume (V):
V = (nRT) / P
Let's substitute the given values into the equation:
n = 0.01 moles
R = 0.0821 L·atm/mol·K
T = 282.15 K
P = 0.973 atm
V = (0.01 moles * 0.0821 L·atm/mol·K * 282.15 K) / 0.973 atm
Simplifying the equation:
V = 2.263 L
Therefore, 0.01 moles of carbon monoxide (CO) at 9 degrees C and 0.973 atm occupies a volume of 2.263 liters.