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Mnenie [13.5K]
3 years ago
6

A flexible balloon contains 0.400 molmol of an unknown polyatomic gas. Initially the balloon containing the gas has a volume of

7.50 m3 and a temperature of 26.0∘C. The gas first expands isobarically until the volume doubles. Then it expands adiabatically until the temperature returns to its initial value. Assume that the gas may be treated as an ideal gas with Cp=33.26J/mol⋅K and γ=4/3.
What is the total heat Q supplied to the gas in the process?
Physics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

Mass released = 8.6 g

Given data:

Initial number of moles nitrogen= 0.950 mol

Initial volume = 25.5 L

Final mass of nitrogen released  = ?

Final volume = 17.3 L

Formula:

V₁/n₁  = V₂/n₂

25.5 L / 0.950 mol = 17.3 L/n₂

n₂ =  17.3 L× 0.950 mol/25.5 L

n₂ = 16.435 L.mol /25.5 L

n₂ = 0.644 mol

Initial mass of nitrogen:

Mass = number of moles × molar mass

Mass = 0.950 mol × 28 g/mol

Mass = 26.6 g

Final mass of nitrogen:

Mass = number of moles × molar mass

Mass = 0.644 mol × 28 g/mol

Mass = 18.0 g

Mass released = initial mass - final mass

Mass released = 26.6 g - 18.0 g

Mass released = 8.6 g

Read more on Brainly.com - brainly.com/question/15623698#readmore

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Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At  the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

Solving  for hmax:

hmax = 1/2 · v² / g

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A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
3 years ago
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