a gas occupies a volume of 410mL at 27 degrees Celcius and 740mm Hg pressure. Calculate the volume the has would occupy at STP.
(P1V1/T1) = (P2V2/T2)
Remember to use T in Kelvin.
P1=98.57kPa
V1=.410L
T1=300.15°K
P2=1atm
T2=273.15°K
V2=?
Use the combine gas law then find the v2,
V2=.3633L
To calculate the volume of the gas at STP (Standard Temperature and Pressure), we need to use the Ideal Gas Law equation:
PV = nRT,
where:
P is the pressure,
V is the volume,
n is the number of moles of gas,
R is the ideal gas constant, and
T is the temperature.
At STP, the temperature is 0 degrees Celsius (273.15 K) and the pressure is 1 atmosphere (atm).
First, let's convert the given information to the correct units:
Volume (V) = 410 mL = 0.410 L (since 1 L = 1000 mL)
Temperature (T) = 27 degrees Celsius = 27 + 273.15 = 300.15 K
Pressure (P) = 740 mm Hg = 740/760 atm (since 1 atm = 760 mm Hg)
Now, we can rearrange the Ideal Gas Law equation to solve for the unknown V (volume at STP):
V₁ / P₁ = V₂ / P₂,
where V₁, P₁ are the initial volume and pressure, and V₂, P₂ are the final volume and pressure at STP.
Substituting the given values into the equation:
0.410 L / (740/760 atm) = V₂ / 1 atm.
Let's solve for V₂:
V₂ = (0.410 L * 1 atm) / (740/760 atm)
V₂ = (0.410 L * 760) / 740
V₂ = 0.4194 L.
Therefore, the volume the gas would occupy at STP is approximately 0.4194 L.
To calculate the volume the gas would occupy at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas
First, let's determine the number of moles of the gas. To do this, we need to use the ideal gas equation and rearrange it to solve for n:
n = PV / RT
Substituting the given values:
P = 740 mmHg (convert to atm by dividing by 760)
V = 410 mL (convert to L by dividing by 1000)
T = 27°C (convert to Kelvin by adding 273.15)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
Now let's solve for n:
n = (740 mmHg / 760 mmHg/atm) * (410 mL / 1000 mL/L) / (0.0821 L·atm/(mol·K) * (27°C + 273.15)
Simplifying:
n = (0.973) * (0.41) / (0.0821 * 300.15)
n ≈ 0.019515 mol
Now that we know the number of moles of gas, we can use it to calculate the volume at STP. At STP, the pressure is 1 atm and the temperature is 0°C (273.15K).
Using the same ideal gas law equation, we rearrange it to solve for V:
V = nRT / P
Substituting the values at STP:
n = 0.019515 mol
R = 0.0821 L·atm/(mol·K)
T = 273.15 K
P = 1 atm
Now let's solve for V:
V = (0.019515 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm)
V ≈ 0.468 L
Therefore, the gas would occupy a volume of approximately 0.468 liters at STP.