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Paul [167]
4 years ago
14

The number density of gas atoms at a certain location in the space above our planet is about 1.05 × 1011 m-3, and the pressure i

s 2.7 × 10-10 Pa in this region.What is the temperature there?
Physics
1 answer:
Anettt [7]4 years ago
7 0

To solve this problem we will apply the concepts given by the ideal gas equation, which mathematically can be described as

PV = NkT

Here

P = Pressure

V = Volume

N = Number of atoms of molecules

k = Bolzmann constant

T = Temperature

Rearranging to find the temperature we have

T = \frac{PV}{Nk}

Since the value given in the exercise is a unit of atoms per volume, we will readjust the equation like this

T = \frac{P}{\frac{N}{V}k}

Replacing we have,

T = \frac{(2.7*10^{-10}N/m^2)}{(1.05*10^{11}/m^3)(1.38*10^{-23}J/K)}

T = 186.3K

T = -86.81\°C

Therefore the temperature is -86.81°C or 186.3K

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A monatomic ideal gas in equilibrium at a pressure of 3 kPa and a temperature of 300 K is initially confined in a volume of 1.3
GuDViN [60]

To solve this problem, it is necessary to apply the concepts related to the change of entropy in function of the Volume in two states due to the number of moles and the ideal gas constant, this can be expressed as

\Delta S = nRln(\frac{V_2}{V_1})

Where,

R = Gas constant

V = Volume (at each state)

At the same time the number of moles of gas would be determined by the ideal gas equation, that is,

n = \frac{PV}{RT}

Where,

P = Pressure

V = Volume

R = Gas Constant

T = Temperature

n = \frac{(1000Pa)(1.3m^3)}{(300K)(8.314Pa\cdot m^3\cdot K^{-1}\cdot mol^{-1})}

n = 0.521mol

Using the value of moles to replace it in the first equation we have

\Delta S = (0.521mol)(8.314Pa\cdot m^3\cdot K^{-1}\cdot mol^{-1})ln(\frac{10.3m^3}{1.3m^3})

\Delta S = 9.01J/K

Therefore the correct option is A.

7 0
4 years ago
A boat moves 60 kilometers east from point A to point B. There, it reverses direction and travels another 45 kilometers toward p
Anton [14]
The displacement is x + (60km - 45km) - x =60km -45 km = 15 km
8 0
3 years ago
A projectile is launched straight up at 135 m/s. Determine how fast it is moving at the top of its trajectory (flight).
suter [353]
<span> It is 0 m/s. At the very top of its flight is right between when it is going up and going down; this is the ephemeral moment when it is perfectly still. </span>
7 0
3 years ago
An athlete starts running from rest such that he covers a distance of 5mm in 0.50s with a constant change in velocity per unit t
gulaghasi [49]

Answer:

50 s

Explanation:

Given:

Δx = 0.005 m

v₀ = 0 m/s

t = 0.50 s

Find: a

Δx = v₀ t + ½ at²

0.005 m = (0 m/s) (0.50 s) + ½ a (0.50 s)²

a = 0.04 m/s²

Given:

Δx = 50 m

v₀ = 0 m/s

a = 0.04 m/s²

Find: t

Δx = v₀ t + ½ at²

50 m = (0 m/s) t + ½ (0.04 m/s²) t²

t = 50 s

6 0
3 years ago
Hallar la distancia que recorre una movil al cabo
Kryger [21]

Answer:

Distancia, S = 136 metros

Explanation:

Dados los siguientes datos;

Aceleración, a = 3 m/s²

Velocidad inicial, u = 5 m/s

Tiempo, t = 8 segundos

Para encontrar la distancia recorrida, usaríamos la segunda ecuación de movimiento;

S = ut + ½at² Sustituyendo en la fórmula, tenemos;

S = 5 × 8 + ½ × 3 × 8²

S = 40 + 1,5 × 64

S = 40 + 96

Distancia, S = 136 metros

3 0
3 years ago
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