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vodomira [7]
3 years ago
6

At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?

Chemistry
1 answer:
algol [13]3 years ago
7 0

Answer:

  • 820 K

Explanation:

As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.

  • KE α T.

Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

  • First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

  • Second, convert 0°C to kelvin:

         T (K) = T(°C) + 273.15 K= 273.15 K

Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.

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A certain liquid X has a normal boiling point of 133.60°C and a boiling point elevation constant Kb= 2.46°C kg mol^-1.Calculate
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Answer:

136.63 °C

Explanation:

ΔTb=Tb solution - Tb pure

Where; Tb pure = 133.60°C

molar mass of solute = 121.14 g/mol

number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles

molality = 0.431 moles/350 * 10^-3 = 1.23 molal

Then;

ΔTb = Kb * m * i

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i = 1

ΔTb = 2.46 * 1.23 * 1

ΔTb = 3.03 °C

Hence;

Tb solution = ΔTb + Tb pure

Tb solution = 3.03 °C + 133.60°C

Tb solution = 136.63 °C

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