What is the Correct Calculation for Pressure in Boyle's Ideal Gas Law Problem?

[rad10k]

Homework Statement

A 3-Litre Vessal gas at a pressure of 200 kPa. 5-Litres of gas at the same pressure if forced into the vessal. Calculate the new Pressure

Homework Equations

P1V1+P2V2

The Attempt at a Solution

P1=200*V1=3 / V2=5 = P2 120 kPa

New Pressure = 120 kPa

I am told this is incorrect has anyone an idea where I am going wrong do I need to add P2 to P1 to get a final pressure of 320 kPa or something?

Thanks

 

[Andrew Mason]

Homework Statement

A 3-Litre Vessal gas at a pressure of 200 kPa. 5-Litres of gas at the same pressure if forced into the vessal. Calculate the new Pressure

Homework Equations

P1V1+P2V2

The Attempt at a Solution

P1=200*V1=3 / V2=5 = P2 120 kPa

New Pressure = 120 kPa

First of all P1V1+P2V2 is not an equation.

Second, you have to be clear about what is happening. What is the final total volume?

Now comes the trick. Assume that the final volume is occupied by the two gases in proportion to the number of molecules of each (how is that related to the original volumes of each?). Use that partial volume of each gas to determine the final pressure of each gas (which, of course, will be the same for each gas). Will it be higher or lower than the original pressure?

AM

 

[rad10k]

The final total volume is 8 L ?

 

[Andrew Mason]

The final total volume is 8 L ?

??. The gas is all in the vessel. So what is the total volume?

AM

 

[gneill]

A 3-Litre Vessal gas at a pressure of 200 kPa. 5-Litres of gas at the same pressure if forced into the vessal. Calculate the new Pressure

The problem doesn't state whether they are the same type of gas, or whether temperature change due to the work done to compress the gas needs to be taken into account (perhaps the compressed gas is allowed to return to its original temperature before the 'pressure reading' is taken).

 

[Andrew Mason]

The problem doesn't state whether they are the same type of gas, or whether temperature change due to the work done to compress the gas needs to be taken into account (perhaps the compressed gas is allowed to return to its original temperature before the 'pressure reading' is taken).

Since we are dealing with Boyles law, we have to assume the final volume is at the same temperature. It doesn't really matter whether it is the same type of gas if that is the case.

AM

 

[RTW69]

You know PV/RT= moles. For the 3 liter case you know P&V, T is a constant so you can find moles (n_3).

Same process for 5 liter case, you know P&V, T is a constant so you can find moles (n_5)

After you mix them together, T is still a constant, volume is 8 liters total moles is n_3 plus n_5. You can find P

 

[Andrew Mason]

You know PV/RT= moles. For the 3 liter case you know P&V, T is a constant so you can find moles (n_3).

Same process for 5 liter case, you know P&V, T is a constant so you can find moles (n_5)

After you mix them together, T is still a constant, volume is 8 liters total moles is n_3 plus n_5. You can find P

The volume is not 8 litres. The gas is all the vessel.

AM

 

[gneill]

Perhaps a clarifying way to look at this is as follows. Imagine that you have an 8L container with a thin partition dividing it into 3L and 5L sections. Both sections of the container are filled with gas at pressure P1.

The partition is then removed so that all the gas is now in one large 8L container. You have 8L of gas at pressure P1. This container is then contracted in size (perhaps one wall is a piston) until its volume is 3L, the same size as the original 3L partition. What's the pressure?

 

[RTW69]

That works or P*V_3=(n_3 + n_5) *R*T solve for P