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IceJOKER [234]
3 years ago
10

When a piston is pushed down on a cylinder, the pressure on the gas in the cylinder increases from 125 kPa to 375 kPa. The initi

al volume is 3.3 L. What is the final volume?
Chemistry
1 answer:
Elden [556K]3 years ago
7 0

Answer:

The final volume is 1.1 L.

Explanation:

Given data:

Initial pressure of gas = 125 KPa

Final pressure of gas = 375 KPa

Initial volume = 3.3 L

Final volume = ?

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

125 Kpa × 3.3 L = 375 Kpa × V₂

V₂ = 412.5 KPa. L/375 Kpa

V₂ = 1.1 L

The volume of gas is reduced to the 1.1 L because of increasing pressure.

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Would you classify the atoms in the table as pure substances?
ivann1987 [24]

Answer:

Yes

Explanation:

Atoms are generally classified as pure substances. A pure substance has the following properties:

  • All parts are the same throughout i.e. homogeneous
  • They have definite composition
  • The cannot easily be broken down into simpler substances by physical means.
  • Separation by physical methods is not easy.
  • They have unique sets of physical properties.

Elements and compounds are generally classified as pure substances. Since atoms are the building blocks of pure substances, they can be classified as one.

3 0
3 years ago
The distribution of Na+ ions across a typical biological membrane is 10 mmol dm-3 inside the cell and 140 mmol dm-3 outside the
trasher [3.6K]

Answer :  The value of \Delta G^o across the membrane is 6.80 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K_{eq}\\\\\Delta G^o=-RT\times \ln (\frac{C_{in}}{C_{out}})

where,

\Delta G^o = standard Gibbs free energy  = ?

R = gas constant = 8.314 J/K.mol

T = temperature = 37^oC=273+37=310K

K_{eq}  = equilibrium constant

C_{in} = concentration inside the cell = 10mmol.dm^{3-}

C_{out} = concentration outside the cell = 140mmol.dm^{3-}

Now put all the given values in the above formula, we get:

\Delta G^o=-RT\times \ln (\frac{C_{in}}{C_{out}})

\Delta G^o=-(8.314J/K.mol)\times (310K)\times \ln (\frac{10mmol.dm^{3-}}{140mmol.dm^{3-}})

\Delta G^o=6.80\times 10^{3}J/mol=6.80kJ/mol

Thus, the value of \Delta G^o across the membrane is 6.80 kJ/mol

8 0
3 years ago
PLS HURRY: 217 mL of Helium gas at 551oC is brought to standard temperature (0oC) while the pressure is held constant. What is t
krok68 [10]
71.9 mL is the answer
6 0
3 years ago
The metric system is only used in France? True or false
Olegator [25]
False it’s used in a lot of other countries
8 0
4 years ago
Read 2 more answers
What determines how much thrust is generated in a solid fuel rocket engine at any given time?
Cerrena [4.2K]

Computer drawing of a solid rocket engine with the equation

for thrust. Thrust equals the exit mass flow rate times exit velocity

plus exit pressure minus free stream pressure times nozzle area.

On this slide, we show a schematic of a solid rocket engine. Solid rocket engines are used on air-to-air and air-to-ground missiles, on model rockets, and as boosters for satellite launchers. In a solid rocket, the fuel and oxidizer are mixed together into a solid propellant which is packed into a solid cylinder. A hole through the cylinder serves as a combustion chamber. When the mixture is ignited, combustion takes place on the surface of the propellant. A flame front is generated which burns into the mixture. The combustion produces great amounts of exhaust gas at high temperature and pressure. The amount of exhaust gas that is produced depends on the area of the flame front and engine designers use a variety of hole shapes to control the change in thrust for a particular engine. The hot exhaust gas is passed through a nozzle which accelerates the flow. Thrust is then produced according to Newton's third law of motion.


The amount of thrust produced by the rocket depends on the design of the nozzle. The smallest cross-sectional area of the nozzle is called the throat of the nozzle. The hot exhaust flow is choked at the throat, which means that the Mach number is equal to 1.0 in the throat and the mass flow rate m dot is determined by the throat area. The area ratio from the throat to the exit Ae sets the exit velocity Ve and the exit pressure pe. You can explore the design and operation of a rocket nozzle with our interactive nozzle simulator program which runs on your browser.


The exit pressure is only equal to free stream pressure at some design condition. We must, therefore, use the longer version of the generalized thrust equation to describe the thrust of the system. If the free stream pressure is given by p0, the thrust F equation becomes:


F = m dot * Ve + (pe - p0) * Ae


Notice that there is no free stream mass times free stream velocity term in the thrust equation because no external air is brought on board. Since the oxidizer is mixed into the propellant, solid rockets can generate thrust in a vacuum where there is no other source of oxygen. That's why a rocket will work in space, where there is no surrounding air, and a gas turbine or propeller will not work. Turbine engines and propellers rely on the atmosphere to provide the oxygen for combustion and as the working fluid in the generation of thrust.


The thrust equation given above works for both liquid and solid rocket engines. There is also an efficiency parameter called the specific impulse which works for both types of rockets and greatly simplifies the performance analysis for rocket engines.



5 0
3 years ago
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