Chemistry, 21.09.2020 07:01 pleasedontspamme

A sample consisting of 1.00 mol Ar is expanded isothermally at 0 degrees C from 22.4 to 44.8 L (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely (against zero external pressure). For the three processes calculate q, w, delta U, and delta H. Assume ideal behavior.

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Answer from: bludov141ox1ocr

a) so for this process q = -w = 1573.25J, ( deltaH = 0 )

b) q = -w = - ( - 1135.68J ) = 1135.68J

c) q = 0

Explanation:

Given that;

Moles of Ar gas is 1.00 mol

isothermal i.e constant temperature expansion of ideal gas deltaU = 0

now since deltaU = 0 = q + w

0 = q + w

q = -w

now for isothermal reversible expansion.

w = -nRTin(Vf/Vt)

= (1.00mol) ( 8.314K^-1.mol^-1) (273K) in(44.8L/22.4L)

= - 1573.25J

so for this process q = -w = 1573.25J

for a perfect gas at constant temperature, deltaH = 0

dH = D( U + PV )

so at constant temperature dU = 0

and PV = constant for perfect gas; d(PV) = 0

THEREfore deltaH = 0

For isothermal i.e constant temperature expansion of ideal gas

deltaU = 0

for a perfect gas at constant temp deltaH = 0

final pressure of the gas in this case is calculated using ideal gas equation

p = nRT/V

we know that {1pa=1Jm^-3}

= [((1.00mol) ( 8.314K^-1.mol^-1) (273K))/(44.8L)] * (1000 Lm^-3)

= 5.07 * 10^4 Pa

for expansion against an external pressure

w = - Pext . deltav

= - (5.07 * 10^4 Pa) ( 44.8L - 22.4L) (1m^3 / 1000L)

= - 1135.68J

q = -w = - ( - 1135.68J ) = 1135.68J

For this scenerio,

deltaU = 0 and deltaH = 0

for free expansion Pext. = 0

therefore Pext * deltaV = 0 * ( 44.8L - 22.4L) = 0

since q = -w, for this process, q = 0

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

hey there!

mass = 6.75 g

volume = 9 cm³

therefore:

d = mass / volume

d = 6.75 / 9

d = 0.7 g/cm³

answer b

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A sample consisting of 1.00 mol Ar is expanded isothermally at 0 degrees C from 22.4 to 44.8 L (a) r...

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