Three moles of nitrogen gas, N2, at 26.8
◦C are
contained in a 35.8 L cylinder.
Find the pressure the gas exerts on the
cylinder walls.
Answer in units of Pa
To find the pressure exerted by the nitrogen gas on the cylinder walls, we can use the ideal gas law:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, we need to convert the temperature from degrees Celsius to Kelvin:
T = 26.8 + 273.15 = 299.95 K
Next, we can plug in the values into the ideal gas law equation:
PV = nRT
P * 35.8 L = 3 moles * (8.314 J/(mol*K)) * 299.95 K
P * 35.8 L = 7478.712 J
Now, we need to convert liters to cubic meters to match the SI unit of Pascal (Pa):
35.8 L = 0.0358 m^3
P * 0.0358 m^3 = 7478.712 J
Finally, we solve for P:
P = 7478.712 J / 0.0358 m^3
P ≈ 208,804.457 Pa
Therefore, the pressure exerted by the nitrogen gas on the cylinder walls is approximately 208,804.457 Pa.
To find the pressure the gas exerts on the cylinder walls, we can use the ideal gas law equation, which states:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
We have the values for the number of moles (n = 3), volume (V = 35.8 L), and temperature (T = 26.8 °C). However, we need to convert the temperature to Kelvin since the ideal gas constant (R) is in units of Kelvin.
To convert Celsius to Kelvin, we use the equation:
T(K) = T(°C) + 273.15
So, T(K) = 26.8 °C + 273.15 = 299.95 K
Now, we can rearrange the ideal gas law equation to solve for pressure (P):
P = (nRT) / V
Plugging in the values:
P = (3 moles * R * 299.95 K) / 35.8 L
The ideal gas constant (R) is approximately 8.314 J/(mol·K).
P = (3 * 8.314 J/(mol·K) * 299.95 K) / 35.8 L
Simplifying the equation, the units of J and moles cancel out, resulting in the pressure in pascals (Pa):
P = (3 * 8.314 * 299.95) / 35.8 Pa
Calculating the expression gives us the pressure in pascals:
P ≈ 62285.26 Pa
Therefore, the pressure the gas exerts on the cylinder walls is approximately 62285.26 Pa.