Question

At what temperature will the average speed of molecules of hydrogen gas doubles the speed of molecules of oxygen gas at 300 K 170 K 75 K 273 K 0 K

Answer 1

Answer:

use the formula of Kelvin to celcius

Answer 2

Concept:

Average speed is defined as the arithmetic mean of the mean of the speeds of different molecules of a gas at a given temperature.

Given:

The average speed of molecules of hydrogen gas doubles the speed of molecules of oxygen gas.

Find: At what temperature will the average speed of molecules of hydrogen gas double the speed of molecules of oxygen gas at 300 K, 170 K, 75 K, 273 K, and 0 K?

Solution:

(a) 300 K

We know that the average speed can be expressed as:

                               Average speed = $\sqrt{} \frac{8RT}{\pi M }$

where R is the universal gas constant.

T is the temperature.

M is the molar mass.

According to the given problem,

Average speed of hydrogen gas = The average speed of oxygen gas

Average speed of hydrogen gas = 2 * average speed of oxygen gas

$\sqrt{} \frac{8RT}{\pi M }$ = 2 * $\sqrt{} \frac{8RT}{\pi M }$

$\sqrt{} \frac{T_{1} }{2 }$ = $\sqrt{} \frac{300 }{32 }$

Squaring both sides and we get,

$\frac{T_{1} }{2 }$ = $\frac{300 }{32 }$

$T_{1}$ = $\frac{300 * 2}{32}$

$T_{1}$ = 18.75 K

Hence, the temperature at which the average speed of molecules of hydrogen gas doubles the speed of molecules of oxygen gas at 300 K is 18.75 K.

(b) 170 K

According to the given problem,

Average speed of hydrogen gas = The average speed of oxygen gas

Average speed of hydrogen gas = 2 * average speed of oxygen gas

$\sqrt{} \frac{8RT}{\pi M }$ = 2 * $\sqrt{} \frac{8RT}{\pi M }$

$\sqrt{} \frac{T_{1} }{2 }$ = $\sqrt{} \frac{170 }{32 }$

Squaring both sides and we get,

$\frac{T_{1} }{2 }$ = $\frac{170 }{32 }$

$T_{1}$ = $\frac{170 * 2}{32}$

$T_{1}$ = 10.625 K

Hence, the temperature at which the average speed of molecules of hydrogen gas doubles the speed of molecules of oxygen gas at 170 K is 10.625 K.

(c) 75 K

According to the given problem,

Average speed of hydrogen gas = The average speed of oxygen gas

Average speed of hydrogen gas = 2 * average speed of oxygen gas

$\sqrt{} \frac{8RT}{\pi M }$ = 2 * $\sqrt{} \frac{8RT}{\pi M }$

$\sqrt{} \frac{T_{1} }{2 }$ = $\sqrt{} \frac{75 }{32 }$

Squaring both sides and we get,

$\frac{T_{1} }{2 }$ = $\frac{75 }{32 }$

$T_{1}$ = $\frac{75 * 2}{32}$

$T_{1}$ = 4.68 K

Hence, the temperature at which the average speed of molecules of hydrogen gas doubles the speed of molecules of oxygen gas at 75 K is 4.68 K.

(d) 273 K

According to the given problem,

Average speed of hydrogen gas = The average speed of oxygen gas

Average speed of hydrogen gas = 2 * average speed of oxygen gas

$\sqrt{} \frac{8RT}{\pi M }$ = 2 * $\sqrt{} \frac{8RT}{\pi M }$

$\sqrt{} \frac{T_{1} }{2 }$ = $\sqrt{} \frac{273 }{32 }$

Squaring both sides and we get,

$\frac{T_{1} }{2 }$ = $\frac{273 }{32 }$

$T_{1}$ = $\frac{273 * 2}{32}$

$T_{1}$ = 17.0625 K

Hence, the temperature at which the average speed of molecules of hydrogen gas doubles the speed of molecules of oxygen gas at 273 K is 17.0625 K.

(e) 0 K

According to the given problem,

Average speed of hydrogen gas = The average speed of oxygen gas

Average speed of hydrogen gas = 2 * average speed of oxygen gas

$\sqrt{} \frac{8RT}{\pi M }$ = 2 * $\sqrt{} \frac{8RT}{\pi M }$

$\sqrt{} \frac{T_{1} }{2 }$ = $\sqrt{} \frac{0 }{32 }$

Squaring both sides and we get,

$\frac{T_{1} }{2 }$ = $\frac{0 }{32 }$

$T_{1}$ = 0 K

Hence, the temperature at which the average speed of molecules of hydrogen gas doubles the speed of molecules of oxygen gas at 0 K is  0 K.

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