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yawa3891 [41]
4 years ago
15

When the temperature of a gas in a rigid container decreases, the particles of the gas move slower and experience fewer collisio

ns with each other and the container walls. Because of this, it is sometimes necessary to add more air to automobile tires in the winter.Which gas law applies in this situation?
Chemistry
2 answers:
4vir4ik [10]4 years ago
6 0
When the temperature of a gas within a rigid container decreases, the particles on average move more slowly and do not collide with one another or the container walls as often. It is the the ideal gas law that applies in this situation and states that a decrease in the temperature of a gas also results in a decrease in the pressure. Thus, automobile tyres which have a colder gas in the winter sometimes need additional air to provide suffiicient pressure. 
notsponge [240]4 years ago
6 0

Answer:

Gay Lusaac's law: The pressure Temperature Law

Explanation:

First off, it is important to understand that the fewer collisions experienced by the particles with each other and the walls of the container is referring to the pressure.

SO, form the question, we can tell that temperature and pressure are directly proportional to each other. As temperature increases, the particles move faster which in turn leads to increased pressure.

After that has been established, we look up the gas laws and find the pone that gives a temperature-pressure relationship and this is the Gay-Lussac's Law: The Pressure Temperature Law. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. As the pressure goes up, the temperature also goes up, and vice-versa.

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A student increases the temperature of a 556 cm3 balloon from 278 K to 308 K. Assuming constant pressure, what should the new vo
Mandarinka [93]
The answer is:  [D]:  " 417 cm³ " .
_____________________________________________________
Explanation:  Use the formula:

V₁ /T₁= V₂ /T₂  ;

in which:  V₁ = initial volume = 556 cm³ ;
                T₁ = initial temperature = 278 K ;
                V₂ = final ("new") temperature = 308 K
                T₂ = final ("new:) volume = ?

Solve for  "V₂" ;

Since:  V₁ /T₁= V₂ /T₂ ;

We can rearrange this "equation/formula" to isolate "V₂" on one side of the equation; and then we can plug in our know values to solve for "V₂" ;
_______________________________________________________
       V₁ /T₁= V₂ /T₂  ;  Multiply EACH side of the equation by "T₂ " :

          →  T₂ (V₁ /T₁) = T₂  (V₂ /T₂) ;
______________________________
to get:

↔ T₂  (V₂ /T₂) = T₂ (V₁ /T₁) ;

     →  V₂ = T₂ (V₁ /T₁) ;
______________________________
Now, plug in our known values, to solve for "V₂" ;
______________________________
    →  V₂ = T₂ (V₁ /T₁) ;
______________________________
    →  V₂ = 308 K ( 556 cm³ /278 K)  ;
             → The units of "K" cancel to "1" ; and we have:
________________________________________________________
    →  V₂ = 308*( 556 cm³ / 278 ) = [(208 * 556) / 278 ] cm³ ;
Note:  We will keep the units of volume as:  "cm³ ";  since all the answer choices given are in units of:  "cm³ " ; {that is, "cubic centimeters"}.

   →  [(208 * 556) / 278 ] cm³ = [ (115,648) / (278) ] cm³ ;
                                        
              → For the "(115,648)" ;  round to "3 (three significant figures)" ;
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              →      (116,000) / (278) = 417.2661870503597122  ;
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                                               → which corresponds with "choice [D]".
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The answer is:  [D]:  "417 cm³ " .
______________________________________________________

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