How to approach a problem involving pressure and volume?

[TyErd]

Homework Statement

I have attached the question. I have no clue with this one. Is there a specific equation that I have to use? How do I approach a problem like this??

Homework Equations

density = mass/volume
specific volume = volume/mass = 1/density
mass flow rate = density x volume flow rate
volume flow rate = cross sectional area x velocity
density of water = 1000kg/m^3
density of air = 1.22521kg/m^3 i think?
total volume = mass x specific volume

The Attempt at a Solution

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[1994Bhaskar]

Write down the ideal gas equation PV=nRT for each container's initially while assuming unknowns.

 

[1994Bhaskar]

Then we know that total no. of moles in system(container 1 + container 2) is constant.
Then write final gas equation PV=(n1+n2)RT.

Sufficient data is given to you.Molar mass of air may be required

 

[TyErd]

okay so the ideal gas law eqs: 500 x 10^3 (1) = n x 287 x (25 +273.15)------container 1.
200 x 10^3 (V) = n x 287 x (35 +273.15)------------container 2.

how do i find Volume knowing that it weighs 5kg. Am i suppose to add the two above equations?? then make it equal to P (V1+V2)=(n1+n2) x 287 x (20 + 273.15) ??

 

[1994Bhaskar]

You know the volume of second tank using ideal gas equation for container-2.You have pressure temperature and number of moles(you need molar mass of air).

 

[1994Bhaskar]

Using this you get V1+v2 in final equation.There you have total no. of moles, total volume and equilibrium temperature.Get equilibrium Pressure

 

[TyErd]

I'm not sure i understand what you mean. Isn't the molar mass used to find R in the ideal gas law equation? I used R as 287 is that right? i don't get how to find n

 

[1994Bhaskar]

molar mass of air is 28.97 and R is 8.31J/mol*K.
You are using wrong values of R.

 

[1994Bhaskar]

In the ideal gas equation for first container you have temperature, pressure, and volume calculate no. of moles for container 1.

 

[1994Bhaskar]

For second container you have mass of air 5kg=5000g.
no. of moles= 5000/28.97=172.59

 

[legendary_]

Here is how to solve the problem:

Use finite states to have a picture of the problem easily.

State 1:
V_a1 = 1 m^3
T_a1 = 25 deg C + 273 = 323 K
P_a1 = 500 kPa
m_b1 = 5 kg
T_b1 = 35 deg C + 273 = 308 K
P_b1 = 200 kPa

State 2:
T_a2 = T_b2 = 20 deg C + 273 = 293 K (thermal equilibrium w/ surroundings)
V_b2 = ?
P_b2 = ?

where a is the first tank, and b is the second tank.

Solution:
1. Obtain unknowns of each tank for easy computations at the initial state or State 1.

1.1 Obtain the mass of the first tank using PV=mRT
m_a1=P_a1*V_a1 / (R*T_a1), where R is 286.9 J/(kgK)

1.2 Obtain the volume of the second tank
V_b1= m_b1 * R * T_b1 / P_b1

2. Analyze the transition from State 1 to State 2, taking note of the parameters in PV=mRT equation.

2.1 When the valve is opened, the volume of the 2 tanks will not obviously change. So the volume of the second tank V_b2 is equal to V_b1.

2.2 Since the valve is already opened, the contents of the 2 tanks are combined, that is: m_a1 + m_b1 = m_a2 = m_b2. Now, we can solve for the final pressure using PV = mRT:
P_b2 = m_b2 * R * T_b2 / V_b2

 

[1994Bhaskar]

How come R is 286.9?It's 8.31 as i said earlier.
You will need molar mass of air.

 

[TyErd]

8.31 I am pretty sure is the universal gas constant. to find the gas constant of air it is universal gas constant divided by molar mass. R of air = universal gas constant (8.31) / molar mass of air(28.956) = 0.2869kPa = 286.9Pa

 

[TyErd]

anyway my working out now is:
container 1: 500 x 10^3 (1) = m x 287 x (25 + 273.15) thus m = 5.84kg.
container 2: 200 x 10^3 (V) = 5 x 287 x (35 + 273.15) thus V = 2.21 m^3

so mass total is 5.84 + 5 = 10.84 kg

therefore P ( 2.21) = 10.84 x 287 x (20 + 273.15), which means P = 412616 Pa = 412.6kPa. Is that correct?

 

[1994Bhaskar]

Then it's okay.However using universal gas constant seems more fundamental.

 

[TyErd]

is my working out correct as well?

 

[1994Bhaskar]

is my working out correct as well?

seems OK.Check for any calculation errors and if your answer matches with correct answer(if known) then fine.

 

[TyErd]

to find the final equilibrium pressure, what volume do i have to use? the volume of the first/second or both added together?

 

[1994Bhaskar]

Both added together.