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vampirchik [111]
3 years ago
11

I have a hot air balloon that has 18.0 g helium gas inside of it. if the pressure is 2.00 atm and the temperature is 297k, what

is the volume of my balloon?
Physics
2 answers:
Anastaziya [24]3 years ago
4 0

M = molar mass of the helium gas = 4.0 g/mol

m = mass of the gas given = 18.0 g

n = number of moles of the gas

number of moles of the gas is given as

n = m/M

n = 18.0/4.0

n = 4.5 moles

P = pressure = 2.00 atm = 2.00 x 101325 Pa = 202650 Pa

V = Volume of balloon = ?

T = temperature = 297 K

R = universal gas constant = 8.314

Using the ideal gas equation

P V = n R T

(202650) V = (4.5) (8.314) (297)

V = 0.055 m³

Sloan [31]3 years ago
3 0

 The  volume  of   balloon is   54.86 L

  <u><em>calculation</em></u>

The volume is calculated using  the ideal  gas equation

That  is PV=nRT

where,

p= 2.00 atm

V= ?

n( number   of  moles)   which  is calculated as  below

moles = mass /molar mass = 18.0g / 4 g/mol = 4.5   moles

R ( gas constant) = 0.0821  L. atm / mol.K

T(Temperature) = 297 K


make  V the subject of the formula by diving  both side by P

V =nRT/P

V= {(4.5 moles x 0.0821 L.atm/mol.K x 297 K) / 2.00 atm} = 54.86 L





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A car driving up a hill at a constant speed experiences no change in its kinetic energy while it's potential energy increases with increasing height, thus none of the options are correct.

Understanding the concept

Consider a car moving up the hill at a constant speed as shown in the figure below. The following forces act on the car:

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Here as the car is moving up the hill at a constant speed, the net force exerted on the car is zero. Also, the kinetic energy of the car will not change as its velocity is constant and the potential energy will change with increasing height. Thus, none of the given options are correct.

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A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potent
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3 years ago
A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f
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Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m​

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ  and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m​

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