Change 50.0 ml at standard conditions to 43 degrees celsius and 750mmHg
Is this a gas? Use (P1V1/T1) = P2V2/T2)
Don't forget T must be kn Kelvin.
To convert 50.0 mL at standard conditions to 43 degrees Celsius and 750 mmHg, we need to use the Ideal Gas Law equation. The Ideal Gas Law equation is as follows:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
We know that the volume is given as 50.0 mL, but we need to convert it to liters. There are 1000 mL in 1 L, so:
V = 50.0 mL / 1000 mL/L = 0.050 L
Now, we need to convert the given temperature of 43 degrees Celsius to Kelvin. To convert Celsius to Kelvin, we use the equation:
T(K) = T(°C) + 273.15
So:
T(K) = 43 °C + 273.15 = 316.15 K
Finally, we convert the given pressure of 750 mmHg to atmospheres. There are 1 atm to 760 mmHg, so:
P = 750 mmHg / 760 mmHg/atm = 0.9868 atm (approximately)
Now, we can substitute the values into the Ideal Gas Law equation and solve for n (the number of moles of gas):
PV = nRT
(0.9868 atm)(0.050 L) = n(0.0821 L·atm/(mol·K))(316.15 K)
Simplifying the equation:
0.04934 = 25.880715n
Divide both sides by 25.880715:
n = 0.001903 mol (rounded to 4 decimal places)
Therefore, 50.0 mL at standard conditions would be equivalent to approximately 0.0019 moles of gas at 43 degrees Celsius and 750 mmHg.