Ideal Gas Law (Thermal Physics)

[blackz]

guys, I'm having difficulty in solving this problem.

Two vessels, one having 3 times volume of the other, are connected through narrow tube of negligible volume. The initial temperature is 290K. The small vessel is then cooled down to 250K while the large vessel is heated up to 400K. What is the final pressure?

Please help me with this. Any pointer will do. Thanks

I tried to use the equation of state (PV=nRT) of both vessel but I could not get the answer. All i get is the ratio between the final mass of small vessel and the final mass of large vessel.

 

[jgens]

I don't think that you have sufficient information to solve the problem because no information about the initial pressure is given. Also, I don't believe that the ideal gas law is the correct law to apply in this case, even if you were to be given the initial pressure. I think that application of either the combined gas law or Gay-Lussac's law would be better. Additionally, the two containers are connected so the pressure in each of the containers should be equal (I think). It's been a while since I've done chemistry or thermal physics so maybe someone else will clear this up.

 

[baba89]

Hi there,

Thank you for reaching out for help with this problem. The Ideal Gas Law (PV=nRT) is definitely the right equation to use in this situation. However, you also need to take into account the fact that the two vessels are connected through a narrow tube.

First, let's define some variables:
P1 = initial pressure in the small vessel
P2 = initial pressure in the large vessel
V1 = volume of the small vessel
V2 = volume of the large vessel
T1 = initial temperature in both vessels
T2 = final temperature in both vessels
n1 = number of moles of gas in the small vessel
n2 = number of moles of gas in the large vessel

Now, we can set up the equation for the small vessel:
P1V1=n1RT1

Since the small vessel is cooled down to 250K, T2 = 250K. We also know that the number of moles of gas remains constant, so n1 = n2. Therefore, we can rewrite the equation as:
P1V1=P2V2 (nRT2 = nRT1)

Now, let's look at the equation for the large vessel:
P2V2=n2RT2

Since the large vessel is heated up to 400K, T2 = 400K. Again, the number of moles of gas remains constant, so n1 = n2. We can rewrite this equation as:
P2V2=P1V1 (nRT2 = nRT1)

Now, we can combine these two equations:
P1V1=P2V2
P2V2=P1V1

We can simplify this to:
P1V1=P1V1
P2V2=P2V2

This shows us that the initial pressure in the small vessel (P1) is equal to the initial pressure in the large vessel (P2). However, since the volume of the large vessel is 3 times that of the small vessel (V2 = 3V1), the initial pressure in the large vessel (P2) will be 1/3 of the initial pressure in the small vessel (P1).

Now, we can use the Ideal Gas Law to find the final pressure in the large vessel:
P2V2=n2RT2
P2(3V1)=n2RT2
P2(