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Luden [163]
3 years ago
11

When the vessel and its contents are warmed to 100 °C, Q decomposes into its constituent elements. What is the total pressure, a

nd what are the partial pressures of xenon and oxygen in the container?

Engineering
1 answer:
Aleksandr [31]3 years ago
4 0

Answer:

The question is incomplete, but here is a similar question with complete details ; Gaseous compound Q contains only xenon and oxygen . When 0.100 g of Q is placed in a 50.0 mL steel vessel at 0 °C, the pressure is 0.229 atm. (a) What is the molar mass of Q (b) When the vessel and its contents are warmed to 100 °C, Q decomposes into its constituent elements . What is the total pressure , and what are the partial pressures of Xenon and Oxygen

Explanation:

The step by step calculations is as shown in the attached file.

It should be noted that what was applied is

  1. The Ideal gas equation PV = nRT
  2. Daltons law of partial pressure which states that in a mixture of gases, the total pressure exerted is equal to the sum of the individual partial pressures of the gases at constant temperature.
  3. It should be noted that the total pressure of the gases can be gotten by applying pressures law at constant volume
  4. P1/P2 = T1/T2
  5. It should also be noted that Partial pressure = Total pressure x Mole fraction

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The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
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