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Sonja [21]
3 years ago
11

Pluto's atmosphere. As recently observed by the New Horizons mission, the surface pressure of Pluto is about 11 microbar. The su

rface temperature is about 37 K.
(a) What is the number density (in units of number per cubic centimeter) of molecules at Pluto's surface (Hint: use ideal gas law)? The radius of Pluto is about 1187 km and the surface gravity is about 0.62 m s. What is the total mass of the atmosphere in terms of Kg?

(b) Calculate the saturation vapor pressure (in units of Pa) of ethane at Pluto's surface. The saturated vapor pressure of ethane can be assumed as: log1o(P)-10.01-1085.0/(T-0.561). T is temperature in K and the vapor pressure (P) in units of millimeters of Hg (~133.32 Pa).

(c) If the volume mixing ratio of ethane on Pluto is about 1%, what is mass mixing ratio of ethane (assume the mean molecular weight is 28 g mol')? What is the partial pressure of ethane at the surface? (Hint: should you use volume mixing ratio or mass mixing ratio to calculate the partial pressure? Think about the physical meaning of gas pressure.) Finally, is ethane condensable at Pluto's surface)

Physics
1 answer:
Komok [63]3 years ago
4 0

Answer:

a) The number density is 3.623 × 10⁻³ \frac{mol}{m^{3} }

The mass of the atmosphere is 1.3 × 10²²Kg

b) The pressure is 10⁻²⁰ Millimeter of mercury

c) The mass mixing ratio is 0.0107

The partial pressure of ethane is 0.01114 Pa

Yes it is condensable because it boiling point is -88.5 C  which is equivalent to 184.5 K i.e is adding 273 to -88.5C and the temperature of the atmosphere  is 37 K.

Explanation:

The explanation is on the first and second uploaded image

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A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and t
Katen [24]

Answer:

0.358g

Explanation:

Density of Helium = 0.179g/L

ρ=m/v

m=ρv

when the volume was 2L

m1= 0.179*2

m1=0.358g

when the volume increased to 4L

m2= 0.179*4

m2=0.716g

gram of helium added = 0.716g-0.358g

=0.358g

5 0
3 years ago
Please help me!! Need this done before the 40 min end
Alex_Xolod [135]

Answer:

its c

Explanation:

bc i know

5 0
3 years ago
A circuit contains two 1.5 volt battery and a bulb with a resistance of 3 ohms. Calculate the current
horrorfan [7]

Answer:

<em>The current is 1 A</em>

Explanation:

<u>Current in a Series Connection </u>

When two or more elements are connected in series, all of them have the same current, and the sum of their individual voltages is the total voltage applied to the circuit.

According to Ohm's law:

V=R.I

Where V is the voltage, R is the resistance and I is the current of a circuit.

We have a voltage of V=1.5 V + 1.5 V = 3 V and a resistance of R=3 ohms.

We can calculate the current by solving for I:

\displaystyle I=\frac{V}{R}=\frac{3}{3}=1\ A

The current is 1 A

3 0
2 years ago
The power of the kettle was 2.6 kW
faltersainse [42]

Answer:

Cp = 4756 [J/kg*°C]

Explanation:

In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.

Q = m*Cp*(T_f - T_i)/t

where:

Q = heat transfer = 2.6 [kW] = 2600[W]

m = mass of the water = 0.8 [kg]

Cp = specific heat of water [J/kg*°C]

T_f  = final temperature of the water = 100 [°C]

T_i = initial temperature of the water = 18 [°C]

t = time = 120 [s]

Now clearing the Cp, we have:

Cp = Q*t/(m*(T_f - T_i))

Now replacing

Cp = (2600*120)/(0.8*(100-18))

Cp = 4756 [J/kg*°C]

7 0
3 years ago
A laser beam is incident on two slits with a separation of 0.230 mm, and a screen is placed 4.75 m from the slits. If the bright
Mariulka [41]

Answer:

7.55\times 10^{-7} m

Explanation:

We are given that

d=0.23 mm=0.23\times 10^{-3} m

1mm=10^{-3} m

Screen is placed  from the slits at distance ,L=4.75 m

The bright interference fringes on the screen are separated  by 1.56 cm.

\Delta y=1.56 cm=1.56\times 10^{-2} m

1 m=100 cm

We have to find the wavelength of laser light.

We know that

\Delta y=\frac{\lambda L}{d}

Substitute the values

1.56\times 10^{-2}=\frac{\lambda\times 4.75}{0.23\times 10^{-3}}

\lambda=\frac{1.56\times 10^{-2}\times 0.23\times 10^{-3}}{4.75}

\lambda=7.55\times 10^{-7} m

4 0
3 years ago
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