calculate the pressure on mmHg of 0.0108mol of methane (CH4) in a 0.265 flask at 37 Celsius
Use PV = nRT. Remember T must be in kelvin. Convert to mm remembering that 1 atm = 760 mm Hg.
To calculate the pressure in mmHg (millimeters of mercury) of 0.0108 moles of methane (CH4) in a 0.265 L flask at 37 degrees Celsius, we will use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in mmHg)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 37 + 273.15
T(K) = 310.15 K
Now we can substitute the given values into the ideal gas law equation and solve for P:
P * 0.265 L = 0.0108 mol * 0.0821 L·atm/(mol·K) * 310.15 K
P * 0.265 L = 0.0108 * 0.0821 * 310.15 L·atm
P * 0.265 = 0.283968 L·atm
P = 0.283968 L·atm / 0.265 L
P ≈ 1.070 mmHg
Therefore, the pressure of 0.0108 moles of methane in a 0.265 L flask at 37 degrees Celsius is approximately 1.070 mmHg.
To calculate the pressure in mmHg of methane (CH4) in a 0.265 L flask at 37 degrees Celsius, we need to use the ideal gas law equation:
PV = nRT
Here, P represents the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin by using the formula:
T(K) = T(°C) + 273.15
T(K) = 37 + 273.15 = 310.15 K
Now, let's substitute the known values into the ideal gas law equation:
P * 0.265 = 0.0108 * 0.0821 * 310.15
P * 0.265 = 0.2676
Divide both sides of the equation by 0.265:
P = 0.2676 / 0.265
P ≈ 1.01 mmHg
Therefore, the pressure of 0.0108 moles of methane in a 0.265 L flask at 37 degrees Celsius is approximately 1.01 mmHg.