If 0.500 mol of nitrogen gas occupies a volume of 11.0 L at 0°C, what volume will 2.15 mol of nitrogen gas occupy at the same temperature and pressure?
Use PV = nRT and solve for P for the first set, then use PV = nRT using P obtained from the first calculation and solve for V.
V = nRT/P .
Note that volume (V) is proportional to the number of moles, n.
So there would be 2.15/0.500 = 4.3 times as much volume.
That would be 47.3 liters.
To solve this problem, we can use the ideal gas law equation: PV = nRT.
First, let's determine the initial pressure (P) and temperature (T):
- The problem states that the initial volume (V) is 11.0 L.
- The problem states that the initial amount of gas (n) is 0.500 mol.
- The problem states that the initial temperature (T) is 0°C. However, we need to convert this to Kelvin. Adding 273.15 to 0°C gives us 273.15 K.
Using the ideal gas law equation, we can rearrange it to solve for the initial pressure (P):
P = (nRT) / V
Substituting the given values:
P = (0.500 mol * R * 273.15 K) / 11.0 L
Next, let's determine the final volume (V) using the given amount of gas (n) and the previously calculated pressure (P) and temperature (T):
- The problem states that the final amount of gas (n) is 2.15 mol.
Rearranging the ideal gas law equation to solve for the final volume (V):
V = (nRT) / P
Substituting the given values:
V = (2.15 mol * R * 273.15 K) / P
Now, we need the value of R, the ideal gas constant. Its value is 0.0821 L·atm/(mol·K).
Substituting the value of R into the equations, we have:
Initial Pressure (P1) = (0.500 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 11.0 L
Final Volume (V2) = (2.15 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / P1
Now, we can calculate the final volume (V2) by substituting the value for the initial pressure (P1) into the equation for the final volume:
V2 = (2.15 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / [(0.500 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 11.0 L]
Simplifying the equation:
V2 = (2.15 mol * 0.0821 L·atm/(mol·K) * 273.15 K * 11.0 L) / (0.500 mol * 0.0821 L·atm/(mol·K) * 273.15 K)
Calculating the final volume (V2):
V2 ≈ 102.51 L
Therefore, 2.15 mol of nitrogen gas will occupy approximately 102.51 L at the same temperature and pressure.
To solve this problem, we can use the ideal gas law, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas.
The ideal gas law equation is given as: PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
To find the volume that 2.15 mol of nitrogen gas will occupy at the same temperature and pressure, we can set up a proportion using the given information.
First, let's convert the temperature from Celsius to Kelvin by adding 273.15:
0°C + 273.15 = 273.15 K
Now, we can set up the proportion based on the number of moles:
0.500 mol / 11.0 L = 2.15 mol / V
Cross-multiplying the equation gives us:
0.500 mol * V = 11.0 L * 2.15 mol
Simplifying further:
V = (11.0 L * 2.15 mol) / 0.500 mol
Calculating the volume:
V = 47.3 L
Therefore, 2.15 mol of nitrogen gas will occupy a volume of 47.3 L at the same temperature and pressure.