Question

Gas is confined in a tank at a pressure of 10.0 atm and a temperature of 15o
C. If
half of the gas is withdrawn and the temperature is raised to 65o
C, what is the new
pressure of the gas?

Answer 1

Pi = Initial Pressure (Atmospheres)
Pf = Final pressure (Atmospheres)
V = Volume
n = Moles
R = That constant number
Ti = Initial Temperature (Kelvins)
Tf = Final Temperature (Kelvins)

Use the Ideal Gas law.
PiV=nRTi & PfV=nRTf
----In the problem, half the gas is removes, so there will be half the moles of gas. If we solve both equations for N, we get:
n=PiV/RTi & n=2PfV/RTf
-----Notice the 2nd equation is multiplied by two in order to keep the same n value as the first equation. This is so we can combine the equations and solve for Pf:
PiV/RTi=2PfV/RTf
----We can cancel out R and V as they keep the same place and value in each equation.
Pi/Ti=2Pf/Tf
----Now let's isolate the value we want to find. Final pressure
Pf=Pi*Tf/Ti*2
----
Plug in the values and solve it. Kelvin must be used.
Pf=(10atm)(338.15K)/(288.15K)(2)
Pf=(3381.5)/(576.3)
Pf=5.867603679...... Atmospheres

Answer 2

you need to multiply by 0.5 not by 2