A 425mL volume of hydrogen chloride gas, HCL, is collected at 25 degrees Celsius and 720 torr. what volume will it occupy?
Am I missing something? It will occupy 425 mL according to the problem.
the question says what will it occupy at STP?
that is what i was confused about?
You left off the at STP part when you posted.
(P1V1/T1) = (P2V2/T2)
To find the volume of hydrogen chloride gas at a different temperature and pressure, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas (in units of force per unit area, such as torr)
V is the volume of the gas (in units of cubic meters or milliliters)
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K) or 62.36 L·torr/(mol·K))
T is the temperature of the gas (in units of Kelvin)
First, let's convert the initial temperature from Celsius to Kelvin by adding 273 to it:
T1 = 25 + 273 = 298 K
Next, we need to calculate the number of moles of hydrogen chloride gas. To do this, we'll use the molar volume of a gas at STP (Standard Temperature and Pressure): 1 mole of any gas occupies 22.4 liters at 0 degrees Celsius (273 K) and 1 atmosphere (760 torr). Since the initial volume is given in milliliters, we need to convert it to liters:
V1 = 425 mL = 0.425 L
Now, we can find the number of moles of hydrogen chloride gas:
n = V1 / (22.4 L/mol) = 0.425 L / 22.4 L/mol = 0.019 mol
Now, let's calculate the final volume of the gas. We'll use the new temperature and pressure values:
T2 = ? (not given)
P2 = 720 torr = 720 torr / 760 torr/atm = 0.947 atm
Now, rearrange the ideal gas law equation to solve for V2:
V2 = (n * R * T2) / P2
Substituting the known values:
V2 = (0.019 mol * 0.0821 L·atm/(mol·K) * T2) / (0.947 atm)
Now we can solve for V2 by dividing both sides by (0.019 mol * 0.0821 L·atm/(mol·K)):
V2 = (T2 * 1 L) / (0.947 atm * 0.019 * 0.0821 L·atm/(mol·K))
Simplifying, we get:
V2 = 54.07 T2
So, the volume of hydrogen chloride gas at the new temperature and pressure will be equal to 54.07 times the new temperature in Kelvin.