a vessel has 6 g oxygen at a pressure P & temp 400 K . a small hole is made in it so that oxygen leaks out . how much oxygen leaks out if the pressure is P/2 & temperature 300 K ?

1. 5g
2. 4g
3. 2g
4. 3g

Use PV = nRT.

There is no pressure listed but make up one to use (the easiest is to use 1 atmosphere). n = grams/molar mass. Solve for volume.
Then use PV = nRT again, this time using P = 1/2 what you used in the first case, change T to the new temperature, and solve for n. Convert n to grams and subtract from 6. Post your work if you get stuck.

I think I worked this problem for you within the last week. Perhaps some other student.

No

To calculate the amount of oxygen that leaks out, we need to apply the ideal gas law. The ideal gas law equation is:

PV = nRT

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

To solve the problem, we can assume that the volume of the vessel remains constant. Therefore, we can write the equation as follows:

P1 * V = n1 * R * T1 ---> Equation 1 (Initial conditions)
(P/2) * V = n2 * R * T2 ---> Equation 2 (Final conditions)

We want to find the difference in the number of moles of gas, which will give us the amount of oxygen that leaks out:

Δn = n1 - n2

First, let's find the number of moles of oxygen under the initial conditions:

P1 * V = n1 * R * T1

Since we are given the pressure P1, temperature T1, and the mass of the oxygen, we can rearrange the equation to solve for n1:

n1 = (P1 * V) / (R * T1)

Next, let's find the number of moles of oxygen under the final conditions:

(P/2) * V = n2 * R * T2

Rearranging the equation to solve for n2:

n2 = (P/2 * V) / (R * T2)

Finally, we can calculate the amount of oxygen that leaks out:

Δn = n1 - n2

Now, let's substitute the values given in the question into the equations and calculate the difference in the number of moles:

Given:
P1 = P
T1 = 400 K
T2 = 300 K

n1 = (P1 * V) / (R * T1)
n2 = (P/2 * V) / (R * T2)
Δn = n1 - n2

Since we are interested in the mass of the oxygen that leaks out, we can calculate it by multiplying the difference in the number of moles of oxygen by the molar mass of oxygen (32 g/mol):

Mass of oxygen leaked = Δn * Molar mass of oxygen

Finally, let's solve the equations to find the answer:

n1 = (P * V) / (R * T1)
n2 = (P/2 * V) / (R * T2)
Δn = n1 - n2
Mass of oxygen leaked = Δn * 32 g