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nignag [31]
3 years ago
7

An ideal gas contained in a piston which is compressed. The gas is insulated so that no heat flows into or out of it. 1) What ha

ppens to the temperature of the gas when it is compressed?
Chemistry
1 answer:
Goryan [66]3 years ago
5 0

Answer:Temperature increases

Explanation: As the gas in the container is an ideal gas so it should follow the ideal gas equation, the equation of state.

We know ideal gas equation to be PV=nRT where

   P=pressure

   V=Volume

   T=Temperature

   R=Real gas constant

   n=Number of moles

since the gas is insulated such that no heat goes into or out of the system .

When we compress the ideal gas using a piston, Thermodynamically it means that work is done on the system by the surroundings.

Now as the ideal gas is been compressed so the volume of the gas would decrease and slowly a time will reach when no more gas can be compressed that is there cannot be any further decrease in volume of the gas.

From the equation PV=nRT

Once there is no further compression is possible hence volume becomes constant so pressure of the ideal gas becomes directly proportional to the temperature as n and R are constants. Also as the pressure and volume are inversely related so an decrease in volume would lead to an increase in pressure.

As the ideal  gas is compressed so the pressure of the gas would increase since the gas molecules have smaller volume available after compression hence the gas molecules would quite frequently have collisions with other gas molecules or piston and this collision would lead to increase in speed of the gas molecules and  so the pressure would increase .

The increase in pressure would lead to an increase in temperature as show by the above ideal gas equation because the pressure and temperature are directly related.

So here we can say that work done on the system by surroundings leads to increase in temperature of the system.

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3 years ago
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Which of the following isoelectronic series is correctly ranked from largest ionic radius to smallest ionic radius? 1. N 3−, O 2
love history [14]

Answer:

<u>Option 1</u>:  N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²    

Explanation:

<u>Ionic radius is the radius of an atom´s ion in ionic crystal structure</u>.<u> </u><u><em>In an ion that lose an electron, to form a cation, the radius of the ion gets smaller</em></u><em>, </em>because the repulsion between electrons decrease because fewer electrons are present. Conversely, <u><em>adding on electron to a neutral atom, to form an anion, causes electron - electron repulsions to increase, so the size of the radius of the ion gets bigger.</em></u>                  

<u><em>Isoelectronic species are ions or elements that have the same number of electrons in their electronic shells but have different overall charges, because of their different atomic numbers</em></u>.                        

<u><em>In a isolelectronic series (same number of electrons),</em></u> <u><em>the increase of the positive charge (given by the number of protons in the nucleus), will cause a decrease in radius </em></u>beacuse the greater electrostatic attraction between the electrons and the nucleus. Consequently, the ion with the greatest nuclear charge will have the smallest ionic radius and the ion with the smallest nulear charge will have the largest ionic radius.  

<u>We will use this principle to solve our problem</u>.  

In our case, the given ions are:  

  • N⁻³ :    Z = 7,  e⁻ = 10
  • O⁻²:     Z= 8,   e⁻ =10
  • F⁻:       Z = 9,  e⁻ = 10
  • Na⁺:    Z= 11,   e⁻ = 10
  • Mg⁺²:  Z=12,   e⁻ =10

where Z= number of protons, and e⁻ = number of electrons.

<em><u>Hence the decreasing order of ionic radius is:</u></em>

N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²  

Have a nice day!

4 0
3 years ago
Why is it better to allow the solvent front to move almost to the top of a paper strip rather than stopping it at the halfway po
m_a_m_a [10]

Answer:

To allow all the elements or compounds to separate complete.

Explanation:

In chromatography, the compounds need some space and time to separate, one from each other, if you  just use the half of the paper strip maybe you will not notice the different spots of compounds. Remember all the substances have different affinity for the solvents, that means, some react very quickly but others need more time as the colors that conform the black color in an ink.

4 0
3 years ago
Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed.Drag the appropriate items to their respective b
Sonja [21]

Answer:

Ag+(aq)+Br−(aq)→AgBr(s)                                NEGATIVE

CaCO3(s)→CaO(s)+CO2(g)2                           POSITIVE

NH3(g)→N2(g)+3H2(g)                                    POSITIVE

2Na(s)+Cl2(g)→2NaCl(s)                                 NEGATIVE

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)           POSITIVE

I2(s)→I2(g)                                                        POSITIVE

Explanation:

We have to remember, to solve this problem, that the entropy of a gas is higher than that of a liquid which in turn  is higher than the solid. Therefore, comparing the reactants and products look for changes in the state of reactants and products. We also have to look for the increase or decrease of moles of each state based on the balanced chemical reaction.

Ag+(aq)+Br−(aq)→AgBr(s)

The reaction product is a single solid and the  the reactants were 2 species in solution. The change in entropy is negative.

CaCO3(s)→CaO(s)+CO2(g)2

Here we have a solid reactant and we have a solid product plus a gas product. The change in entropy is positive.

NH3(g)→N2(g)+3H2(g)

We have 4 mole gases as products starting from 1 mol reactant gas, the entropy has increased.

2Na(s)+Cl2(g)→2NaCl(s)

In this reaction 2 mol solid Na and 1 mol Cl₂ gas are converted into 2 mol solid NaCl, the entropy has decreased.

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)

The products are 7 mol of gas versus 6 mol of gas reactants and therefore entropy has increased.

I2(s)→I2(g)

1 mol solid I₂ goes into 1 mol gas making the change in  the entropy higher.

4 0
3 years ago
The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves
adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

<em>Where ΔT is change in temperature = 0.400°C</em>

<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>

<em>m is molality of the solution (Moles of solute / kg of solvent)</em>

<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

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6 0
3 years ago
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