What the velocity from the graph given above?

3.27 moles of an ideal gas in a 50.0 L tank has a pressure of 171000 Pa. What is the temperature of the gas? (Unit=degrees C)

laiz [17]

The temperature of the gas is 41.3 °C.

Answer:

The temperature of the gas is 41.3 °C.

Explanation:

So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923× $10^{-6}$ atm. Thus, 171000 Pa = 1.6876 atm.

We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.

Then substituting all the values in ideal gas equation ,we get

1.6876×50=3.27×0.0821×T

Temperature = $\frac{84.38}{0.268467} =314.3 K$

So the temperature is obtained to be 314.3 K. As 0°C = 273 K,

Then 314.3 K = 314.3-273 °C=41.3 °C.

Thus, the temperature is 41.3 °C.

3 0

4 years ago

Calculate the angular velocity of the neptune (orbital period of 165 earth period) in its orbit around the sun.

ludmilkaskok [199]

The correct answer is: Angular velocity = $1.208 * 10^{-9}$ rad/s

Explanation:
The angular velocity is given as:
ω = $\frac{2\pi}{T}$ --- (1)

Where T = 165 * (365 days) * (24 hours/day) * (60 minutes/hour) * (60 seconds/minute) = 5203440000 s

Plug in the value in (1):
ω = $\frac{2\pi}{5203440000} = 1.208 * 10^{-9}$ rad/s

7 0

3 years ago

An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's a

vladimir1956 [14]

I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s

8 0

3 years ago

Read 2 more answers

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

Galina-37 [17]

Answer:

220 A

Explanation:

The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.

So, F = BI₁L

F = (μ₀I₂/2πd)I₁L

F = μ₀I₁I₂L/2πd

Given that the current in the rods are the same, I₁ = I₂ = I

So,

F = μ₀I²L/2πd

Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²

So, F = W

μ₀I²L/2πd = mg

making I subject of the formula, we have

I² = 2πdmg/μ₀L

I = √(2πdmg/μ₀L)