Calculate the volume at s.t.p. occupied by a gas 'Q' originally occupying 153.7cc at 287K and 750mm pressure [vapor pressure
of gas 'Q'at 287K is 12mm of Hg]
I understand the question but not the parenthetical part.
Ignoring the part in parentheses, use (P1V1/T1) = (P2V2/T2)
note: I understand vapor pressure to be the pressure of the gas above and in equilibrium with a liquid. If gas Q has a pressure of 750 mm Hg at 287 K I don't understand how it can have a pressure of 12 mm Hg at 287 K.
To calculate the volume of gas 'Q' at standard temperature and pressure (STP), you 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)
In this case, we need to convert the given values to the appropriate units. Let's start with the pressure:
Given pressure = 750 mmHg
Vapor pressure of gas 'Q' at 287K = 12 mmHg
To obtain the pressure of gas 'Q' at 287K above the atmospheric pressure, we subtract the vapor pressure from the given pressure:
Pressure of the gas above atmospheric pressure = Given pressure - Vapor pressure
= (750 mmHg - 12 mmHg)
= 738 mmHg
Next, we convert the pressure from mmHg to atm:
Pressure of the gas above atmospheric pressure = 738 mmHg * (1 atm / 760 mmHg)
≈ 0.9705 atm
Now, let's work on the temperature:
Given temperature = 287 K
Next, we convert the volume from cc to liters:
Given volume = 153.7 cc
Volume at STP = 153.7 cc * (1 L / 1000 cc)
= 0.1537 L
Now, we can use the ideal gas law to find the number of moles:
PV = nRT
n = (PV) / (RT)
= (0.9705 atm * 0.1537 L) / (0.0821 L·atm/(mol·K) * 287 K)
≈ 0.0072 mol
Finally, we can calculate the volume of gas 'Q' at STP using the number of moles:
V = (nRT) / P
= (0.0072 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
≈ 0.162 L
Therefore, the volume of gas 'Q' at STP is approximately 0.162 liters.