given: V1=410 ml, T1=27 degrees celsius, P1= 740mm. what is the V2 in ml at STP?
(P1V1/T1) = (P2V2/T2)
Solve for V2. Don't forget T must be in kelvin.
363
To find the volume at STP (Standard Temperature and Pressure) given the initial volume (V1), temperature (T1), and pressure (P1), we will need to use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15:
T1 = 27 + 273.15 = 300.15 K
Now, we can rearrange the ideal gas law equation to solve for V2 (the final volume at STP):
(P1)(V1) / (T1) = (P2)(V2) / (T2)
Considering STP conditions:
P2 = 1 atm
T2 = 273.15 K
Substituting the values into the equation:
(740 mm)(410 ml) / (300.15 K) = (1 atm)(V2) / (273.15 K)
Now, we can solve for V2:
V2 = [(740 mm)(410 ml) / (300.15 K)] * (273.15 K) / (1 atm)
Let's calculate this value:
V2 ≈ 372.77 ml
Therefore, the volume at STP (V2) is approximately 372.77 ml.