Physics, 11.03.2020 06:10 vperez719

On a microscopic level, the entropy of a system increases with the number of available microstates, which are the possible arrangements of the positions and kinetic energies of the particles in the system. These states cannot be examined individually but are instead studied by using statistical analysis. Each thermodynamic system has a unique number of microstates associated with it, designated as W. Entropy increases with the number of microstates

Consider a system consisting of a cylinder with a movable piston containing 106 gas molecules at 298 K at a volume of 1 L. Consider the following descriptions of this system:

A) Initial system, as stated in the problem statement.

B) Starting from the initial system, the volume of the container is changed to 2 L and the temperature to 395 K.

C) Starting from the initial system, a combination reaction occurs at constant volume and temperature.

D) Starting from the initial system, the gas reacts completely to produce 107 gas molecules at 395 K in a volume of 2 L.

E) Starting from the initial system, the temperature is changed to 395 K at constant volume.

Arrange the descriptions in order of increasing number of microstates in the resulting system.

Rank from most to fewest microstates. To rank items as equivalent, overlap them

Answers: 2

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Answers

Answer from: Sruyle51230

The items are arranged as follows:

A) Initial system, as stated in the problem statement.

D) Starting from the initial system, the gas reacts completely to produce 107 gas molecules at 395 K in a volume of 2 L.

C) Starting from the initial system, a combination reaction occurs at constant volume and temperature.

E) Starting from the initial system, the temperature is changed to 395 K at constant volume.

B) Starting from the initial system, the volume of the container is changed to 2 L and the temperature to 395 K.

Explanation:

The increase in the number of molecules results in the increase in the number of microstates. This results in the possible arrangements of the molecules. Apart from that, there is an increase in the volume of the container as the increase of the volume increases the number of positions where individual molecules can be found. Thus, the order above shows how the microstates increase.

Answer from: Quest

inverse v=i*r

nice nice

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge