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ZanzabumX [31]
2 years ago
12

Explain why doesn’t the total pressure increase when more gas is added to the chamber?

Chemistry
1 answer:
german2 years ago
4 0

Pressure is inversely proportional to volume. Therefore, the effect of pressure change is opposite to the effect of volume change. So when more gas is added to the chamber the total pressure of the chamber doesn't increase.

<h3>What are the different relations between pressure and volume?</h3>
  • As the volume changes, the concentrations and partial pressures of both reactants and products change.
  • As the volume decreases, the reaction shifts to the reaction side with fewer gas particles.
  • As the volume increases, the reaction shifts to the side of the reaction containing more gas particles.
  • As the pressure increases, the equilibrium shifts towards reactions with fewer moles of gas.
  • As the pressure decreases, the equilibrium shifts to the side of the reaction with higher moles of gas.
  • Moreover, the pressure change in the system due to the addition of the inert gas is not limited to this.

To know more about Pressure and Volume visit:
brainly.com/question/5018408

#SPJ4

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A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
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a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

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