A McLeod gauge measures low gas pressures by compressing a known volume of the gas at constant temperature. If 315 cm^3 of gas is compressed to a volume of 0.0457 cm^3 under a pressure of 2.51 kPa what was the original gas pressure?
P1V1 = P2V2.
To solve this problem, we can use the ideal gas law, which states that the product of pressure (P), volume (V), and temperature (T) is constant for a given amount of gas.
The equation for the ideal gas law is:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
In this case, we can assume that the temperature is constant. Therefore, we can rewrite the equation as:
P1V1 = P2V2
Where:
P1 = original pressure
V1 = original volume
P2 = final pressure
V2 = final volume
Let's substitute the given values into the equation:
P1 * 315 cm^3 = 2.51 kPa * 0.0457 cm^3
To make the units consistent, we need to convert cm^3 to liters and kPa to Pa:
P1 * 315 * 10^(-3) L = 2.51 * 10^3 Pa * 0.0457 * 10^(-3) L
Simplifying the units:
P1 * 0.315 L = 2.51 * 0.0457 J
Dividing both sides by 0.315 L:
P1 = (2.51 * 0.0457) / 0.315
Calculating the right side of the equation:
P1 = 0.1144 J / 0.315
P1 ≈ 0.3636 J
Therefore, the original gas pressure was approximately 0.3636 J.
To find the original gas pressure, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure (in pascals)
V is the volume (in cubic meters)
n is the number of moles
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature (in kelvin)
In this case, we want to find the original gas pressure, so we can rearrange the equation as follows:
P1V1 = P2V2
Where:
P1 is the original gas pressure
V1 is the original volume (315 cm^3)
P2 is the final gas pressure (2.51 kPa)
V2 is the final volume (0.0457 cm^3)
First, we need to convert the volumes to cubic meters:
V1 = 315 cm^3 = 3.15 x 10^-4 m^3
V2 = 0.0457 cm^3 = 4.57 x 10^-8 m^3
Next, we need to convert the final pressure to pascals:
P2 = 2.51 kPa = 2.51 x 10^3 Pa
Now, we can plug these values into the equation:
P1 * 3.15 x 10^-4 m^3 = 2.51 x 10^3 Pa * 4.57 x 10^-8 m^3
Simplifying and solving for P1:
P1 = (2.51 x 10^3 Pa * 4.57 x 10^-8 m^3) / (3.15 x 10^-4 m^3)
P1 ≈ 0.00367 Pa
Therefore, the original gas pressure was approximately 0.00367 Pa.