Answer:
5.5 × 10^14 Hz
Explanation:
The wavelength can be represented using the formula;
λ = v/f
Where;
λ = wavelength (m)
v = speed of light (3 × 10^8m/s)
f = frequency (Hz)
According to this question, a light associated with the visible spectrum has a wavelength of 550 nm.
550nm = 550 × 10^-9m
f = v/λ
f = 3 × 10^8 ÷ 550 × 10^-9
f = 0.0055 × 10^ (8+9)
f = 0.0055 × 10^17
f = 5.5 × 10^14 Hz
The frequency is 5.5 × 10^14 Hz
Answer:
The outside temperature is -45.8°C
Explanation:
When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).
V1 / T1 = V2 / T2
2.95L/298K = 2.25L / T2
(2.95L/298K ) . T2 = 2.25L
T2 = 2.25L . 298K / 2.95L
T2 = 227.2K
T°K - 273 = T°C
227.2K - 273 = -45.8°C
Explanation:
Entropy means the amount of randomness present within the molecules of the body of a substance.
Relation between entropy and microstate is as follows.
S = 
where, S = entropy
= Boltzmann constant
= number of microstates
This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.
Also, let us assume that
= 1, and
= 0.833
Therefore, change in entropy will be calculated as follows.

= 
= 
= 
or, = 
Thus, we can conclude that the entropy change for a particle in the given system is
J/K particle.
<u>Answer:</u> The boiling point of water in Tibet is 69.9°C
<u>Explanation:</u>
To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg (Conversion factor: 1 atm = 760 mmHg)
= final pressure = 240. mmHg
= Heat of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature or normal boiling point of water = ![100^oC=[100+273]K=373K](https://tex.z-dn.net/?f=100%5EoC%3D%5B100%2B273%5DK%3D373K)
= final temperature = ?
Putting values in above equation, we get:
![\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B240%7D%7B760%7D%29%3D%5Cfrac%7B40700J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B373%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D%5C%5C%5C%5C-1.153%3D4895.36%5B%5Cfrac%7BT_2-373%7D%7B373T_2%7D%5D%5C%5C%5C%5CT_2%3D342.9K)
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:


Hence, the boiling point of water in Tibet is 69.9°C