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1 year ago

How many mL will a 0.205 mole sample of He occupy at 3.00 atm and 200 K? Report your answer to the nearest mL.

Chemistry

1 answer:

Tcecarenko [31]1 year ago

4 0

1.1214 mL will a 0.205-mole sample of He occupy at 3.00 atm and 200 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Using equation PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 3.00 atm

V= ?

n=0.205 mole

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$

T=200 K

Putting value in the given equation:

$\frac{nRT}{P} =V$

$V= \frac{0.205 \;mole\;0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 200}{3 \;atm}$

V= 1.1214 mL

Learn more about the ideal gas here:

brainly.com/question/27691721

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where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

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R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $785.0K$

Now put all the given values in this formula, we get:

$\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]$

$K_2=1.45\times 10^{-2}s^{-1}$

Therefore, the rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

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