Question

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A truck tire has, for all practical purposes, constant volume. Suppose the pressure is 120 kPa in the morning when the temperature is 30. degrees C. In the afternoon, the tire temperature is 55 degrees C. What will the tire pressure be at that time?

Answer 1

Answer:

196 is the answer pls make me brainlist

Answer 2

Concept Used -

=> The concept of gas laws is used here -

    $\dfrac{P1*V1}{T1} = \dfrac{P2*V2}{T2}$

   Where -

   P1 - initial pressure of a sample of gas

   V1 = initial volume of a gas sample

   T1 = Initial temperature

   P2 = Final pressure

   V2 = Final volume

   T2 = Final temperature

=> The gas laws have been derived by establishing a relation between the Boyle's law and the Charles Law which state that -

$V\propto \dfrac{1}{P}\ (Boyle's\: Law\:-At\:Constant\:Temperature)\\\\\\V\propto T\ (Charles\:Law\:-\:At\:Constant\:Pressure)$

=> Here, the volume of gas in the tire remains same only that the pressure and the temperature vary.

Given -

=> Air temperature in the tire rises from 30 degrees C in the morning to 55 degrees C in the afternoon.

=> The volume of the air in the tire remains same.

To find -

=> The Tire pressure in the afternoon.

Listing the values according to the gas equation -

P1 = 120 kPa (kilopascals)    P2 = ?

V1 = x litres (let)                    V2 = x litres (As the volume remains same)

T1 = 30 °C                             T2 = 55 °C

   => 30 + 273 K                        => 55 + 273 K

T1 = 303 K                            T2 = 328 K (°C to K)

Putting the values in the Gas Equation, we get-

➼ $\dfrac{120 * x}{303} =\dfrac{P2 * x}{328}$

➼ $\dfrac{120}{303} = \dfrac{P2}{328}$  [Cancelling x from both L.H.S and R.H.S]

Now we have,

➼ $\dfrac{120 * 328}{303} = P2$ [Multiplying both sides by 328]

➼ $\dfrac{39360}{303} =P2$

➼ 129.9 = P2

➼ P2 = 129.9 kPa

Answer -

=> The tire pressure in the afternoon will be 129.9 kPa or kilopascals .

More to know -

1) Relation between Pressure and Temperature for a gas -

=> We already know about the Boyle's Law and Charles Law -

$V\propto \dfrac{1}{P}\ (Boyle's\: Law\:-At\:Constant\:Temperature)\\\\\\V\propto T\ (Charles\:Law\:-\:At\:Constant\:Pressure)$

=> From this, it is clear that -

$V \propto \dfrac{T}{P}$ [Combining both of them)

=> Now, taking P (Pressure) to L.H.S we get,

$V * P\propto T$

=> If we remove Volume from here (Assuming Volume as a constant) -

$P \propto T\:[At\:Constant\:Volume]$

Conclusion - At a constant volume, the pressure of a gas varies directly with its temperature. This is the concept used in the sum. This is generally the case for gases enclosed in a rigid body that cannot change its shape.

2) Concept of Liquefaction of gases -

=> In the liquefaction of gases, gases are subjected to immense amounts of pressure, due to this the temperature of the gas decreases and the gas liquefies (eg. Ammonia and Oxygen are liquefied by this method)

=> But doesn't this contradict the relation we proved earlier ?!

No, it doesn't, because during liquefaction, the gases are pressurized in such a manner that they are free to change their volume , now, they obey the Boyle's law and their volume decreases, the following take place -

• On subjecting gases to very high pressure and slowly letting them expand during liquefaction -

• The gas molecules are brought close together and volume decreases.
• Due to this, the kinetic energy of the gas molecules decreases.
• Due to decrease in the average kinetic energy of the molecules, the temperature reduces and the gases liquefy.
• This cannot be done at normal atmospheric conditions.