Question

On planet x, the absolute pressure at a depth of 2.00 m below the surface of a liquid nitrogen lake is 5.00 × 105 n/m2. at a depth 5.00 m below the surface, the absolute pressure is 8.00 × 105 n/m2. the density of liquid nitrogen is 808 kg/m3. (a) what is the atmospheric pressure on planet x? (b) what is the acceleration due to gravity on planet x?

Answer 1

Pressure inside the liquid nitrogen is calculated by Pascal's law

it is given by the equation

`P = P_o + \rho g h`

given that

at 2 m below the surface pressure is
`P = 5 * 10^5 Pa`

so this equation is given as

`5 * 10^5 = P_o + 2\rho*g`

also we have
`P = 8 * 10^5 Pa` at 5 m below

`8 * 10^5 = P_o + 5\rho*g`

in order to find gravity at the planet subtract two equations

`(8 - 5)*10^5 = 3 *\rho* g`

`3* 10^5 = 3 * 808 * g`

`g = 123.76 m/s^2`

now to find the atmospheric pressure we can use any one equation

`P_o = 5 * 10^5 - 2* 808 * 123.76`

`P_o = 3 * 10^5 Pa`

Answer 2

a)
`(3.00).10^(5)Pa`

b)
`123.762(m)/(s^(2))`

Step-by-step explanation:

Let's start defining the following unit of pressure :

`1Pascal=1Pa=1(N)/(m^(2))`

Where N is newton and m is meter.

Within a mass of liquid and defining the surface of the liquid as depth 0.00 m, we can calculate the absolute pressure in any point of the liquid as :

`Pabs=Pgauge+Patm`

Where Pgauge is the pressure because of the height of liquid over the point.

Where Patm is the atmospheric pressure.

The gauge pressure in a point of a mass of liquid is define as :

Pgauge = δ.g.h

Where δ is the density of the liquid.

Where g is the acceleration due to gravity.

Where h is the height of liquid over the point.

Using the data from the exercise :

`Pabs=Pgauge+Patm`

At a depth of 2.00 m :

`(5.00).10^(5)(N)/(m^(2))=808(kg)/(m^(3)).g.(2.00)m+Patm` (I)

At a depth of 5.00 m :

`(8.00).10^5(N)/(m^(2))=808(kg)/(m^(3)).g.(5.00)m+Patm` (II)

In (II) :

`Patm=(8.00)10^(5)(N)/(m^(2))-808(kg)/(m^(3)).g.(5.00m)` (III)

Putting (III) in (I) :

`(5.00).10^(5)(N)/(m^(2))=808(kg)/(m^(3)).g.(2.00)m+(8.00).10^(5)(N)/(m^(2))-808(kg)/(m^(3)).g.(5.00)m`

`-(3.00).10^(5)(N)/(m^(2))=1616(kg)/(m^(2)).g-4040(kg)/(m^(2)).g`

`-(3.00).10^(5)(N)/(m^(2))=-2424(kg)/(m^(2)).g`

`g=((3.00).10^(5))/(2424)(N.m^(2))/(m^(2).kg)`

`g=123.762(m)/(s^(2))`

This is the gravity for point b)

For a) we need to replace this value in another equation. For example in equation (III) :

`(8.00).10^(5)(N)/(m^(2))-808(kg)/(m^(3)).(123.762(m)/(s^(2))).5m=Patm`

`Patm=(3.00).10^(5)(N)/(m^(2))`

`Patm=(3.00).10^(5)Pa`

And that is the value of atmospheric pressure on planet x.