A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.
The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:
The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.
Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.
The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, 
A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
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Answer:
Explanation:
It is easier if you convert the kelvin temperature into Celsius degrees:
- ºC = T - 273.15 = 150 - 273.15 = -123.15ºC
Now, you know that that is a very cold temperature. Thus, may be the oxygen is not gas any more but it changed to liquid . . . or solid?
You must search for the boiling point and melting (freezing) point of oxygen in tables or the internet. At standard pressure (about 1 atm) they are:
- Melting point: −218.79 °C,
- Boiling point: −182.962 °C
That means that:
- below -218.79ºC oxygen is solid (not our case).
- between -218.79ºC and -182.962ºC oxygen is liquid (not our case)
- over -182.962ºC oxygen is a gas. This is our case, because -123.15ºC is a higher temperature than -182.962ºC.
Hence, <em>the state of matter of oxygen at 150K</em>, and standard pressure, is gas.
Answer:
Initially the function is symmetric with respect to the axis of the one dimensional box. In the final state it is also symmetrical, however you can envision a snapshot of the system as the light field is interacting with the wave-function wherein a node begins to develop as is shown in the middle and the wave function is evolving from the initial to final state. Now consider that the electron density during process is the square of the wave function:
Electron density during transition
As can be seen in the initial and final states the electron density is symmetrically distributed with respect to the axis of the box. However with the field on, the electron density is not symmetrically distributed and a transitory dipole moment can be present. To relate back to real molecules think of each of those orbitals as a linear combination of atomic orbitals. One important factor is the symmetry. But there may be one other factor that will be just as important as symmetry. If you treat orbital 1 as a linear combination over n orbitals and orbital 2 as a linear combinations of orbitals as well, there will be a spatial over lap between the orbital in the ground state and the orbital in the excited state. If there is no spatial overlap between the ground state and excited state orbitals there will be no transition dipole moment. However, if the electrons are in the same place spatially, a large transition dipole moment will result.
Explanation:
