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yulyashka [42]
4 years ago
6

What is a change that will not affect the pressure in a container

Physics
2 answers:
Dimas [21]4 years ago
7 0

Answer:

changing the material that the fluids container is made of.

Lubov Fominskaja [6]4 years ago
7 0

<u>A change that will not affect the pressure in a container:</u>

The factors that affect the pressure inside a container are PVT (Pressure, Volume and Temperature). Apart from these three factors, changing or tweaking any other factor like, changing the material using which the container was made, etc would not affect the pressure.

In this case of a container, a change in pressure(adding or removing air from the container), volume (changing the amount of fluid present in the container), temperature (changing the temperature in which the container is kept) would all affect the pressure inside the container while something like, changing the material using which the container was made, would not change the pressure.

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A cube of copper was found to have a mass of 630 g. What are the dimensions of the cube? (The density of copper is 8.94 g/cm3.)​
raketka [301]

Answer:

Density = mass/volume

Volume of cube = mass/density

630g/8.94 = 70.5cm^3 (3s.f.)

Since it is a cube, its dimensions of Length x breadth x height=volume

Take 70.5 amd cube root it

\sqrt[3]{70.4697986577)

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3 0
3 years ago
An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum
Tju [1.3M]

Answer:

The pressure is  P_2  =  4.25 \ a.t.m

Explanation:

From the question we are told that

   The initial pressure is P_1  =  11.2\ a.t.m

   The  temperature is  T_1 =   299 \ K

   

Let the first volume be  V_1 Then the final volume will be  2 V_1

 Generally for a diatomic  gas

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Here r is the radius of the molecules which is  mathematically represented as

    r =  \frac{C_p}{C_v}

Where C_p \  and\   C_v are the molar specific heat of a gas at constant pressure and  the molar specific heat of a gas at constant volume with values

     C_p=7 \  and\   C_v=5

=>   r =  \frac{7}{5}

=>  11.2*( V_1 ^{\frac{7}{5} } ) =  P_2  *  (2 V_1 ^{\frac{7}{5} } )

=>   P_2  =  [\frac{1}{2} ]^{\frac{7}{5} } * 11.2

=>  P_2  =  4.25 \ a.t.m

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Answer:

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