A pocket of gas is discovered in a deep drilling operation. The gas has a temperature of 480c and is at a pressure of 12.8 atm. What volume of gas is required to provide 18.0 L of gas at the surface where the conditions are 22 C and 1.00 atm?
Use PV = nRT and solve for n at the "pocket" conditions. Then PV = nRT again and substitute the surface conditions but use n from the first calculation.
Alternatively, you may use P1V1/T1 = P2V2/T2
To solve this problem, we can use the ideal gas law equation, which states:
PV = nRT
Where:
- P is the pressure
- V is the volume
- n is the number of gas moles
- R is the ideal gas constant (0.0821 L*atm/(mol*K))
- T is the temperature in Kelvin
First, let's convert the temperatures to Kelvin.
The given temperature of the gas at the drilling operation is 480°C. To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature:
T1 = 480°C + 273.15 = 753.15 K
The surface temperature is given as 22°C:
T2 = 22°C + 273.15 = 295.15 K
Next, we can rearrange the ideal gas law equation to solve for the volume (V2) at the surface:
V2 = (n1 * R * T2) / P2
Where:
- V2 is the volume at the surface (we need to find this)
- n1 is the number of gas moles (we'll assume it remains constant)
- R is the ideal gas constant
- T2 is the temperature at the surface
- P2 is the pressure at the surface (given as 1.00 atm)
To find n1, we'll use the ideal gas law equation for the initial conditions at the drilling operation:
PV = nRT1
n1 = (PV) / (RT1)
Substituting the given values:
n1 = (12.8 atm * V1) / (0.0821 L*atm/(mol*K) * 753.15 K)
Now, we have all the values we need to solve for V2:
V2 = ((12.8 atm * V1) / (0.0821 L*atm/(mol*K) * 753.15 K)) * (0.0821 L*atm/(mol*K) * 295.15 K) / 1.00 atm
Substituting V1 = 18.0 L:
V2 = ((12.8 atm * 18.0 L) / (0.0821 L*atm/(mol*K) * 753.15 K)) * (0.0821 L*atm/(mol*K) * 295.15 K) / 1.00 atm
Simplifying this expression will give us the final volume (V2).