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3 years ago

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The root-mean-square speed (thermal speed) of the molecules of a gas is 200m/s at 23.0°C. At 227°C the root-mean-square speed (t

hermal speed) of the molecules will be closest to (1) 160 m/s. (2) 330 m/s. (3) 260 m/s. (4) 630 m/s. (5) None of the above.

1 answer:

777dan777 [17]3 years ago

8 0

Answer:

330 m/s approx

Explanation:

The RMS speed of a gas is proportional to square root of its absolute temperature is

V ( RMS ) ∝ √T

$\frac{V_1}{V_2} =\sqrt{\frac{T_1}{T_2} }$

Here V₁ = 200 , T₁ = 23 +273 = 300K , T₂ = 227 +273 = 500 K

Putting the values

200 / V₂ = $\sqrt{\frac{300}{500} }$

V₂ = 330 m/s approx

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This question is incomplete, the missing image uploaded along this answer below;

Answer:

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Answer:

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Explanation:

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The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

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$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

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Calculate the energy required to achieve that temperature change:

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2 years ago

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Answer:

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Explanation:

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