Response:
(a) 
(b) 
Clarification:
Greetings.
(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:
This value is negative as it expands against the given pressure.
(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:
The moles are calculated based on the provided mass of argon:
Consequently, the work amounts to:
Best regards.
The light's wavelength absorbed during the transition is 459 nm. Energy difference between the 5-d and the 6-s sub-levels in gold is expressed as ΔE. Let the wavelength associated with the electron's transition from the 5-d to the 6-s state be λ. The relationship that describes the connection between energy and wavelength is defined as: E = hc/λ, where E stands for photon energy, h represents Planck's constant, c is the speed of light, and λ denotes the wavelength of the photon. Therefore, the absorption wavelength in this transition stands at 459 nm.
Answer:
b = 0.6487 kg / s
Explanation:
In the context of oscillatory motion, friction is related to velocity,
fr = - b v
where b represents the friction coefficient.
Upon solving the equation, the angular velocity is represented as
w² = k / m - (b / 2m)²
In this case, we're given an angular frequency w = 1Hz, the mass m = 0.1 kg, and the spring constant k = 5 N / m. This allows us to derive the friction coefficient.
Let’s denote
w₀² = k / m
w² = w₀² - b² / 4m²
b² = (w₀² -w²) 4 m²
Now, let's calculate the angular frequencies.
w₀² = 5 / 0.1
w₀² = 50
w = 2π f
w = 2π 1
w = 6.2832 rad / s
Substituting values yields
b² = (50 - 6.2832²) 4 0.1²
b = √ 0.42086
b = 0.6487 kg / s
