Aweather balloon is inflated to a volume of 29.1 l at a pressure of 733 mmhg and a temperature of 24.7 ∘c. the balloon rises in the atmosphere to an altitude where the pressure is 365 mmhg and the temperature is -14.5 ∘c. assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

volume balloon  5.06 10⁻² m³

Explanation:

The balloon is inflated with air, which in the first instance behaves like an ideal gas, which has as an equation

    PV = n R T

Where R is the ideal gas constant that is worth 8.314 J / mol K.

Let's use the initial data to calculate the number of moles of gas in the balloon that remains constant, but first we reduce all quantities to the SI system

    V1 = 29.1 L (1 m2 / 1000 L) = 29.1 10⁻³ m³

    P1 = 733 mmHg (1 105 Pa / 760 mmHg) = 0.964 10⁵ Pa

    T1 = 24.7 + 273.15 = 297.85K

    P2 = 365 mmHg (1 105 Pa / 760 mmHg) = 0.480 10⁵ Pa

    T2 = -14.5 +273.15 = 258.65K

Let's clear and calculate the number of moles

    n = P V / R T

    n = P1 V1 / R T1

    n = 0.964 10 5 29.1 10⁻³ / (8.314 297.85)

    n = 1.13 moles

With the data of the second part we clear the volume and calculate

   V = n R T / P

   V2 = n R T2 / P2

   V2 = 1.13 8.314 258.65 /0.480 10⁵

   V2 = 5.062 10⁻² m³

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.