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AlekseyPX
2 years ago
14

10. When the pressure on a gas inetcases three times, by how much will the volume incrcase or decrease?

Chemistry
1 answer:
blagie [28]2 years ago
4 0

Answer:The answer to this question comes from experiments done by the scientist Robert Boyle in an effort to improve air pumps. In the 1600's, Boyle measured the volumes of gases at different pressures. Boyle found that when the pressure of gas at a constant temperature is increased, the volume of the gas decreases. when the pressure of gas is decreased, the volume increases. this relationship between pressure and volume is called Boyle's law.

Explanation: So, at constant temperature, the answer to your answer is: the volume decreases in the same ratio as the ratio of pressure increases.

BUT, in general, there is not a single answer to your question. It depend by the context.

For example, if you put the gas in a rigid steel tank (volume is constant), you can heat the gas, so provoking a pressure increase. But you won't get any change in volume.

Or, if you heat the gas in a partially elastic vessel (as a tire or a soccer ball) you will get both an increase of volume AND an increase of pressure.

FINALLY if you inflate a bubblegum ball, the volume will be increased without any change in pressure and temperature, because you have increased the NUMBER of molecules in the balloon.

There are many other ways to change volume and pressure of a gas that are different from the Boyle experiment.

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2 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

brainly.com/question/20305871

#SPJ1

5 0
1 year ago
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