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Masteriza [31]
3 years ago
5

Which of the following will have the slowest rate of diffusion at a given temperature?

Chemistry
1 answer:
marysya [2.9K]3 years ago
8 0
<h3>Answer:</h3>

Chlorine gas (Cl₂)

<h3>Explanation:</h3>
  • According to the Graham's law of diffusion, the diffusion rate of a gas is inversely proportional to the square root of its density or molar mass.
  • Therefore, a lighter gas will diffuse faster at a given temperature compared to a heavy gas.
  • Consequently, the heavier a gas is then the denser it is and the slower it diffuses at a given temperature and vice versa.

In this case we are given gases, CI₂

, H₂,He and Ne.

  • We are required to identify the gas that will diffuse at the slowest rate.
  • In other words we are required to determine the heaviest gas.

Looking at the molar mass of the gases given;

Cl₂- 70.91 g/mol

H₂- 2.02 g/mol

He - 4.00 g/mol

Ne- 20.18 g/mol

Therefore, chlorine gas is the heaviest and thus will diffuse at the slowest rate among the choices given.

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if the Celsius temperature of a gas at constant pressure is increased from 10 Celsius to 20 Celsius the volume is
svet-max [94.6K]

Answer:

               The volume is increased.

Explanation:

                     According to <em>Charles' Law</em>, " <em>at constant pressure the volume and temperature of the gas are directly proportional to each other</em>". Mathematically this law is presented as;

                                                    V₁ / T₁ = V₂ / T₂   -----(1)

In statement the data given is,

T₁  =  10 °C  =  283.15 K                    ∴  K  =  273.15 + °C

T₂  =  20 °C  =  293.15 K

So, it is clear that the temperature is being increased hence, we will find an increase in volume. Let us assume that the starting volume is 100 L, so,

V₁  =  100 L

V₂  =  Unknown

Now, we will arrange equation 1 for V₂ as,

                                                    V₂  =  V₁ × T₂ / T₁

Putting values,

                                     V₂  = 100 L × 293.15 K / 283.15 K

                                     V₂  = 103.52 L

Hence, it is proved that by increasing temperature from 10 °C to 20 °C resulted in the increase of Volume from 100 L to 103.52 L.

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