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34kurt
3 years ago
6

If the absolute temperature of a gas is tripled, what happens to the root‑mean‑square speed of the molecules?

Chemistry
1 answer:
Gnesinka [82]3 years ago
4 0

Answer:

D. The new rms speed is 1.732 times the original rms speed.

Explanation:

The expression for the root mean square speed is:

C_{rms}=\sqrt {\dfrac {3RT}{M}}

R is Gas constant having value = 8.314 J / K mol  

M is the molar mass of gas

T is the absolute temperature

As seen from the formula, root mean square speed is directly proportional to the square root of the absolute temperature.

So,

C_{rms}\propto \sqrt {T}

Given, absolute temperature of a gas is tripled, so, the new rms speed will be √3 (1.732) of the original.

<u>Hence, the correct option is:- D. The new rms speed is 1.732 times the original rms speed.</u>

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\boxed{\text{4 m $\cdot$ s$^{-2}$}}

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4 0
3 years ago
If the detector is capturing 3.3×108 photons per second at this wavelength, what is the total energy of the photons detected in
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Answer:

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

Explanation:

The energy carried by electromagnetic radiation is displaced by waves. This energy is not continuous, but is transmitted grouped into small "quanta" of energy called photons. The energy (E) carried by electromagnetic radiation can be measured in Joules (J). Frequency (ν or f) is the number of times a wave oscillates in one second and is measured in cycles / second or hertz (Hz). The frequency is directly proportional to the energy carried by a radiation, according to the equation: E = h.f, (where h is the Planck constant = 6.63 · 10⁻³⁴ J / s).

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. it is expressed in units of length (m). In light and other electromagnetic waves that propagate at the speed of light (c), the frequency would be equal to the speed of light (≈ 3 × 10⁸ m / s) between the wavelength :

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E=5.93*10^{-23} \frac{J}{proton} *3.3*10^{8} \frac{proton}{s} *\frac{60}{1} \frac{s}{minute} *\frac{60}{1} \frac{minute}{hr}

E=7.04*10⁻¹¹ \frac{J}{hr}

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

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