A gas occupies a volume of 256mL at a pressure of 250torr at 32 degrees Celsius. Calculate the pressure (in atmosphere)if the volume is reduced to 156mL and the temp is increased to 62 degrees Celsius. I really need the formula that I would be using. I am confused because using PV=nRT there is no moles being given here.
I surely would appreciate it. THanks!
Two ways to do it.
Use PV = nRT for the first set of conditions and solve for n, then redo PV = nRT, use the n from above and the new conditions, and solve for pressure.
Second way is to use
(P1V1/T1) = (P2V2/T2)
The second way is a little shorter since you cut out the solving for n step.
Thanks! I knew the formula but wasn't quite sure if i convert mL to L in the formula. I know that I must covert C to K and Torr to Atm for the equation.
.628 atm
To solve this problem, you can use the combined gas law equation, which relates the initial and final conditions of pressure, volume, and temperature of a gas sample. The formula is:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Where:
P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure (which we need to calculate)
V₂ = final volume (156 mL in this case)
T₂ = final temperature (62 degrees Celsius in this case)
Since you are given the temperature in Celsius, you will need to convert it to Kelvin by adding 273.15 to the Celsius temperature.
Now, let's substitute the given values into the formula:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Substituting the values:
(250 torr * 256 mL) / (32 + 273.15 K) = (P₂ * 156 mL) / (62 + 273.15 K)
To calculate P₂, rearrange the formula to solve for P₂:
P₂ = [(250 torr * 256 mL) / (32 + 273.15 K)] * [(62 + 273.15 K) / 156 mL]
Now, plug in the numbers and calculate:
P₂ = [(250 torr * 256 mL) / (32 + 273.15 K)] * [(62 + 273.15 K) / 156 mL]
P₂ ≈ 1.62 atm
Therefore, the pressure in atmospheres (atm) when the volume is reduced to 156 mL and the temperature is increased to 62 degrees Celsius is approximately 1.62 atm.