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Alchen [17]
3 years ago
12

Thermodynamic Properties: Two identical, sealed, and well-insulated jars contain different gases at the same temperature. Each c

ontainer contains the same number of moles of gas. Container 1 contains helium, a monotomic gas with a molecular weight of 4.0 kg/kMol. Container 2 contains CO2, a triatomic linear molecule with a molecular weight of 44 kg/kMol. (a) Which gas has higher internal energy? (b) Which gas has higher translational energy? (c) Which gas has higher pressure?
Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

Explanation:

Let us assume that the gas is ideal gas in the given problem.

(a)  Hence, expression for internal energy of a monoatomic gas is as follows.

            U = \frac{3}{2}RT

As there are three kinds of translations possible for a mono atomic ideal gas. Therefore, no rotation is possible.

And, according to equipartition theorem, each possible rotation, translation or vibration can contribute  to the internal energy of the system.

And, for Helium  a mono atomic ideal gas,

     U_{He} = \frac{3}{2}RT

For carbon dioxide, which is considered a linear triatomic molecule, there are  3 translations possible, 2 rotations possible and 4 vibrations possible.

But vibrations contribute RT to the energy

So, U_{CO_{2}} = \frac{3}{2}RT + \frac{2}{2}RT + 4RT = \frac{13}{2}RT

Therefore,  has higher internal energy.

(b) Irrespective of the type of the molecule, there are only 3 translation states possible. So, translational kinetic energy is equal to

               U_{trans} = \frac{3}{2}RT

is equal for both helium and carbon dioxide.

(c)  It is known that the ideal gas equation is given as follows.

                  PV = nRT

Here, it is given that

T is same (same temperature)

n is same (no of moles of gas)

V is same (identical container)

R is a constant.

So, P is equal for both gases.

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A bird flies 3.7 meters in 46 seconds, what is its speed?
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Answer:

Speed is 0.08 m/s.

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1. A 4000-kg truck traveling with a velocity of 20 m/s due south collides headon with a 1350-kg car traveling with a velocity of
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(a) The momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) The size of momentum is 93500 kgm/s and it will be directed towards South.

Explanation:

The mass of the truck moving due south is given as 4000 kg and the speed is 20 m/s. Similarly, the mass of the car moving due north is 1350 kg and the speed is 10 m/s.

(a) Then the momentum of each vehicle can be obtained by the product of mass with their respective speed.

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Similarly, the momentum of car will be

Momentum of car = 1350 * 10 = 13500 kgm/s

So, the momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) Since, after collision, the vehicles stick together, the momentum after collision will be equal to the total momentum of both the vehicles before collision. This is because, it will obey conservation of momentum.

Momentum of vehicles after collision = total momentum before collision

Momentum after collision = 80000+13500 = 93500 kgm/s.

The direction of the vehicles after collision will be towards south as the mass and speed of the truck is greater than car. So the impact or force exerted on the car by the truck will be greater and thus both the vehicles will be directed towards south after collision.

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