1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Serjik [45]
3 years ago
5

A tank has a volume of 0.1 m3 and is filled with He gas at a pressure of 5 x 106 Pa. A second tank has a volume of 0.15 m' and i

s filled with He gas at a pressure of 6 x 106 Pa. A valve connecting the two tanks is opened. Assumıng He to be a monatomic ideal gas and the walls of the tanks to be adiabatic and rigıd, find the final pressure of the system Hint: Note that the internal energy is constant.
Chemistry
1 answer:
GarryVolchara [31]3 years ago
5 0

Answer:

P=5.6*10^{6} Pa

Explanation:

Consider that, as the system is adiabatic, U_{1}= U_{2} where U1 and U2 are the internal energies before the process and after that respectively.

Consider that: U=H-PV, and that the internal energy of the first state is the sum of the internal energy of each tank.

So, H_{1}-P_{1}V_{1}=H_{2}-P_{2}V_{2}\\H_{1}^{A} -P_{1}^{A}V_{1}^{A}+H_{1}^{B} -P_{1}^{B}V_{1}^{B}=H_{2}-P_{2}V_{2}

Where A y B are the tanks. The enthalpy for an ideal gas is only function of the temperature, as the internal energy is too; so it is possible to assume: H_{1}=H_{2}\\H_{1}^{A}+H_{1}^{B} =H_{2}

So, P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}=P_{2}V_{2}

Isolating P_{2},

P_{2}=\frac{P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}}{V_{2}}

V_{2}=V_{1}^{A}+V_{1}^{B}=0.25m^{3}

So,

P_{2}=\frac{5000000*0.1+6000000*0.15}{0.25}=5600000Pa=5.6*10^{6} Pa

You might be interested in
Consider the five balanced chemical reactions listed below, all using O2 as a reactant. Normally, O2 is an excess reagent for re
ad-work [718]

Answer: option E. None because in all the reactions O2 is in excess

Explanation:

7 0
3 years ago
The alveoli are surrounded by __________ carrying blood to and from the heart.
natka813 [3]
The alveoli are surrounded<span> by tiny blood vessels, called capillaries. The </span>alveoli<span> and capillaries both have very thin walls, which allow the oxygen to pass from the </span>alveoli<span>to the blood. The capillaries then connect to larger blood vessels, called veins, which bring the oxygenated blood from the lungs to the heart.</span>
4 0
3 years ago
Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric acid that contains 0.500 mol of HCl are combined and react to c
jok3333 [9.3K]

Answer:

2.24 L of hydrogen gas, measured at STP, are produced.

Explanation:

Given, Moles of magnesium metal, Mg = 0.100 mol

Moles of hydrochloric acid, HCl = 0.500 mol

According to the reaction shown below:-

Mg_{(s)} + 2HCl_{(aq)}\rightarrow MgCl_2_{(aq)} + H_2_{(g)}

1 mole of Mg reacts with 2 moles of HCl

0.100 mol of Mg reacts with 2*0.100 mol of HCl

Moles of HCl must react = 0.200 mol

Available moles of HCl = 0.500 moles

Limiting reagent is the one which is present in small amount. Thus, Mg is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of Mg on reaction forms 1 mole of H_2

0.100 mole of Mg on reaction forms 0.100 mole of H_2

Mole of H_2 = 0.100 mol

At STP,  

Pressure = 1 atm  

Temperature = 273.15 K

Volume = ?

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K  

<u>⇒V = 2.24 L</u>

2.24 L of hydrogen gas, measured at STP, are produced.

4 0
3 years ago
What does solubility mean?
jasenka [17]

Answer: the ability to be dissolved, especially in water.

Explanation: I think the answer you've picked is right

Hope this helps

8 0
3 years ago
At what temperature is the following reaction feasible: HCl(g) + NH3(g) -&gt; NH4Cl(s)?
Nutka1998 [239]
Energy is distributed not just in translational KE, but also in rotation, vibration and also distributed in electronic energy levels (if input great enough, bond breaks).

All four forms of energy are quantised and the quanta ‘gap’ differences increases from trans. KE ==> electronic.

Entropy (S) and energy distribution: The energy is distributed amongst the energy levels in the particles to maximise their entropy.

Entropy is a measure of both the way the particles are arranged AND the ways the quanta of energy can be arranged.

We can apply ΔSθsys/surr/tot ideas to chemical changes to test feasibility of a reaction:

ΔSθtot = ΔSθsys +  ΔSθsurr

ΔSθtot must be >=0 for a chemical change to be feasible.

For example: CaCO3(s) ==> CaO(s) + CO2(g) 

ΔSθsys = ΣSθproducts – ΣSθreactants 

ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) 

ΔSθsurr is –ΔHθ/T(K) and ΔH is very endothermic (very +ve),

Now ΔSθsys is approximately constant with temperature and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall.

But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800oCΔSθtot becomes plus overall (and ΔGθ becomes negative), so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.

CaCO3(s) ==> CaO(s) + CO2(g)  ΔHθ = +179 kJ mol–1  (very endothermic)

This important industrial reaction for converting limestone (calcium carbonate) to lime (calcium oxide) has to be performed at high temperatures in a specially designed limekiln – which these days, basically consists of a huge rotating angled ceramic lined steel tube in which a mixture of limestone plus coal/coke/oil/gas? is fed in at one end and lime collected at the lower end. The mixture is ignited and excess air blasted through to burn the coal/coke and maintain a high operating temperature.
ΔSθsys = ΣSθproducts – ΣSθreactants
ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) = (40.0) + (214.0) – (92.9) = +161.0 J mol–1 K–1
ΔSθsurr is –ΔHθ/T = –(179000/T)
ΔSθtot = ΔSθsys +  ΔSθsurr
ΔSθtot = (+161) + (–179000/T) = 161 – 179000/T
If we then substitute various values of T (in Kelvin) you can calculate when the reaction becomes feasible.
For T = 298K (room temperature)

ΔSθtot = 161 – 179000/298 = –439.7 J mol–1 K–1, no good, negative entropy change

For T = 500K (fairly high temperature for an industrial process)

ΔSθtot = 161 – 179000/500 = –197.0, still no good

For T = 1200K (limekiln temperature)

ΔSθtot = 161 – 179000/1200 = +11.8 J mol–1 K–1, definitely feasible, overall positive entropy change

Now assuming ΔSθsys is approximately constant with temperature change and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall. But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800–900oC ΔSθtot becomes plus overall, so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.
You can approach the problem in another more efficient way by solving the total entropy expression for T at the point when the total entropy change is zero. At this point calcium carbonate, calcium oxide and carbon dioxide are at equilibrium.
ΔSθtot–equilib = 0 = 161 – 179000/T, 179000/T = 161, T = 179000/161 = 1112 K

This means that 1112 K is the minimum temperature to get an economic yield. Well at first sight anyway. In fact because the carbon dioxide is swept away in the flue gases so an equilibrium is never truly attained so limestone continues to decompose even at lower temperatures.

8 0
3 years ago
Read 2 more answers
Other questions:
  • In a chemical reaction, the mass of the products ____.
    7·2 answers
  • Can an organism be made of only one cell?
    11·1 answer
  • How many molecules are present in 1.75 moles of dihydrogen monoxide
    15·2 answers
  • Special substances in cells that absorb or release hydrogen ions to keep the ph value constant are called
    8·1 answer
  • What happens to kinetic energy when you increade the speed​
    6·2 answers
  • Which of the following represents a compound made of five molecules? CO 5 C 2O 5 C 5O 5CO 2
    8·1 answer
  • Explain the difference between a front and an air mass.
    7·1 answer
  • Question 6
    12·1 answer
  • Please help me! <br><br> please explain your work
    8·1 answer
  • Some elements produce distinct colors in a flame. what is the cause of this phenomenon?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!