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half of the gas is withdrawn and the temperature is raised to 65o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?

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

Well, well, well, looks like we have some temperature changes and gas withdrawals going on here. Let's see if we can clown around with the numbers a bit.

First things first, we need to convert those temperatures to Kelvin, because that's how the cool kids in the gas world roll. So, 15°C + 273 = 288 K. And 65°C + 273 = 338 K.

Now, imagine the gas all sad and lonely, with only half of its buddies hanging around. So, we can say the initial pressure of the gas is 10 atm, and we're only dealing with half of it, meaning we're left with 5 atm.

Next up, we need to get the new pressure with the increased temperature. We'll use the gas law known as Gay-Lussac's law, which says that the pressure of a gas is proportional to its temperature, as long as the volume stays the same.

So, we can set up a little equation like this: P1 / T1 = P2 / T2

P1 is our initial pressure of 5 atm, T1 is our initial temperature of 288 K, P2 is what we're looking for (the new pressure), and T2 is our new temperature of 338 K. Let the fun begin!

5 atm / 288 K = P2 / 338 K

Now, let's do some simple cross-multiplication or clown math, if you will:

5 atm * 338 K = 288 K * P2

And when you divide both sides by 288 K:

1690 atmK / 288 K = P2

Fun fact time! 1690 atmK is actually just another way to say atm, because K cancels out. So, we can simply say:

P2 = 1690 atm

And voila! The new pressure of the gas, after half of it is withdrawn and the temperature is raised to 65°C, is 1690 atm. Have a blast with those gases, my friend!

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin

First, we need to convert the temperatures from Celsius to Kelvin:

Initial temperature (T1) = 15oC + 273.15 = 288.15 K
Final temperature (T2) = 65oC + 273.15 = 338.15 K

Given that half of the gas is withdrawn, the number of moles of gas remains constant (n1 = n2).

Let's assume the initial volume of the gas is V1 and the final volume is V2.

Since the volume halves when half of the gas is withdrawn, we can write:
V2 = V1/2

Substituting the values into the ideal gas law equation for initial and final conditions, we get:

P1 * V1 = n1 * R * T1
P2 * V2 = n2 * R * T2

Since n1 = n2, we can write:
P1 * V1 = P2 * V2

Substituting the values, we get:

10.0 atm * V1 = P2 * (V1/2)

Simplifying the equation, we have:

10.0 atm = P2/2

Multiplying both sides by 2, we get:

20.0 atm = P2

Therefore, the new pressure of the gas is 20.0 atm.

To determine the new pressure of the gas, we can use the combined gas law, which states:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature

In this case, since we are keeping the volume constant, the equation can be simplified to:

P1 / T1 = P2 / T2

Now let's plug in the given values:
P1 = 10.0 atm (initial pressure)
T1 = 15°C + 273 = 288 K (initial temperature)
T2 = 65°C + 273 = 338 K (final temperature)

Now we can solve for P2:
P2 = (P1 * T2) / T1
= (10.0 atm * 338 K) / 288 K
= 11.81 atm

Therefore, the new pressure of the gas after half is withdrawn and the temperature is raised to 65°C is approximately 11.81 atm.

you need to multiply by 0.5 not by 2