The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. cylinder 50 liters contains oxygen at a temperature of 200 K and a pressure of 6 atmospheres , what would the gas pressure be if the volume was changed to 25 liters and the temperature raised to 250 K?

Given that we know the relation between volume, pressure and temperature (given for the ideal gas), we want to find the change in pressure associated to the given change in volume and temperature.

We will see that the pressure after the change is 15 atm.

Let's see how to solve this problem:

So we start by knowing that the volume depends directly on the temperature and inversely on the pressure, then the relation is something like:

V = k*(T/P)

Where k is a constant.

We know that a cylinder of 50lt is at a temperature of 200K and with a pressure of 6 atm, replacing that in the above equation we get:

50lt = k*(200K/6atm)

With this we can find the value of k:

50lt*(6atm/200K) = k = 1.5 lt*atm/K

Then the equation is just:

V = (1.5 lt*atm/K)*(T/P)

Now the problem says that the volume is 25lt and the temperature is 250K, replacing that in the equation we get:

25lt =  (1.5 lt*atm/K)*(250K/P)

Solving for P we have:

P = (1.5 lt*atm/K)*(250K/25lt) = 15atm

Concluding, the pressure is after the change is 15 atm

If you want to learn more about ideal gases, you can read:

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15 atm

Step-by-step explanation:

V varies directly with T and inversely with P.

V = kT/P

When T = 200 and P = 6, V = 50.

50 = k (200) / (6)

k = 1.5

When V = 25 and T = 250:

25 = 1.5 (250) / P

P = 15

a rotation about point l

step-by-step explanation:

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