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katen-ka-za [31]
3 years ago
14

A balloon behaves so that the pressure isP=C2V1/3 and C2 = 100 kPa/m. The balloon is blown up with air from a starting volume of

1 m3 to a volume of 4 m3. Find the final mass of air, assuming it is at 25◦C, and the work done by the air.
Physics
1 answer:
rewona [7]3 years ago
8 0

Explanation:

As it is given that,

            P_{1} = C_{2}V^{\frac{1}{3}}

           P_{1}V^{-\frac{1}{3}}_{1} = C_{2}

As the system is not gaining or losing heat. So, it is an adiabatic process in an assumed ideal gas. The polytropic extent n is \frac{-1}{3}.

            P_{1} = C_{1}V\frac{1}{3}_{1}

                       = (100)(1)^{\frac{1}{3}}

                        = 100 kpa

         P_{2} = C_{2}V^{\frac{1}{3}}_{2}

                      = 100(4)^{\frac{1}{3}}

                      = 158.74 kpa

Now, work done by the air is as follows.

            W_{2} = \int PdV

                       = \frac{P_{2}V_{2} - P_{1}V_{1}}{1 - n}

                       = \frac{158.74 \times 4 - 100 \times 1}{1 - (\frac{-1}{3})}

                       = 401.22 kJ

Work done by the air is as follows.

           P_{2}V_{2} = m_{2}RT_{2}

               m_{2} = \frac{P_{2}V_{2}}{RT_{2}}

                           = \frac{158.74 \times 4}{0.287 \times (273 + 25)}

                           = 7.424 kg

Thus, we can conclude that final mass of air is 7.424 kg and work done by the air is 401.22 kJ.

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I need help with this question how to solve it for Brass and Cooper
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Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

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An atom with seven valence electrons would most likely lose an electron to become stable true or fasle
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Answer:

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Magnetic flux=\phi=BA=B(\pi r^2)

Circumference,C=2\pi r

r=\frac{C}{2\pi}

r=\frac{168}{2\pi} cm

\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s

\int dr=-\int \frac{15}{2\pi}dt

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r=-\frac{15}{2\pi}t+\frac{168}{2\pi}

E=-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}

E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}

t=8 s

B=0.9

E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})

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Answer:

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