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uranmaximum [27]
3 years ago
7

Si una masa dada de hidrógeno ocupa 40 litros a 700 grados torr. ¿Qué volumen ocupará a 1 atmósfera de presión? (dar la presión

en atmósferas). Por favor :c
Chemistry
2 answers:
Svetach [21]3 years ago
7 0

Answer:

V_2=36.84L

Explanation:

Hola,

En este caso, podemos usar la ley de Boyle, la cual nos permite analizar el comportamiento volumen-presión en un gas ideal de manera inversamente propocional:

P_1V_1=P_2V_2

Así, dado el volumen y la presión inicial, la cual se convierte a atmósferas (760 torr = 1atm), calculamos el volumen final a 1 atm como se muestra a continuación:

V_2=\frac{P_1V_1}{P_2}=\frac{700torr*\frac{1atm}{760torr}*40l }{1atm}\\  \\V_2=36.84L

Saludos!

MissTica3 years ago
5 0

Answer:

V_2=~36.84~Litros

Explanation:

En este caso tenemos diferentes valores de volumen y presión para un <u>primer estado</u>:

V_1=~40~Litros

P_1=~700~torr

Y tenemos valores de presión para un <u>segundo estado</u>:

V_2=~?

P_2=~1~atm

Necesitamos una relación entre volumen y presión para poder hallar nuestra incognita ( V_2). Nuestra relación es la <u>ley de Boyle</u>:

P_1*V_1=P_2*V_2

Ahora bien, tenemos una unidad diferente para el presión en P_2 (atm), debemos hacer una <u>conversión</u>. Esta conversión puede hacerse si sabemos que <u>1 atm = 760 torr</u> por lo cual P_2=760~torr

Ahora podemos <u>reemplazar en la ecuación</u>:

700~torr*40~Litros=760~torr*V_2

V_2=\frac{700~torr*40~Litros}{760~torr}

V_2=~36.84~Litros

El volumen sera de <u>36.84 Litros</u>

Espero que ayude!

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(a) Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of t
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Answer:

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The radius of an impurity atom occupying FCC octahedral site is 0.414{\rm{R}}

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The radius of an impurity atom occupying FCC tetrahedral site is 0.225{\rm{R}} .

Explanation:

In order to get a better understanding of the solution we need to understand that the concept used to solve this question is based on the voids present in a unit cell. Looking at the fundamentals

An impurity atom in a unit cell occupies the void spaces. In FCC type of structure, there are two types of voids present. First, an octahedral void is a hole created when six spheres touch each other usually placed at the body center. On the other hand, a tetrahedral void is generated when four spheres touch each other and is placed along the body diagonal.

Step 1 of 2

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The position of an atom that fits in the octahedral site with radius \left( r \right)is as shown in the first uploaded image.

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The radius of the impurity is as follows:

2r=a-2R------(A)

The relation between radius of atom and edge length is calculated using Pythagoras Theorem is shown as follows:

Consider \Delta {\rm{XYZ}} as follows:

(XY)^ 2 =(YZ) ^2 +(XZ)^2

Substitute XY as{\rm{R}} + 2{\rm{R + R}} and {\rm{YZ}} as a and {\rm{ZX}} as a in above equation as follows:

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The radius of an impurity atom occupying FCC octahedral site is 0.414{\rm{R}}

Note

An impure atom occupies the octahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The relation between the edge length and radius of atom is calculated using Pythagoras Theorem. This further enables in finding the radius of an impure atom.  

Step 2 of 2

(2)

The impure atom in FCC tetrahedral site is present at the body diagonal.

The position of an atom that fits in the octahedral site with radius rr is shown on the second uploaded image :

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The body diagonal is represented by AD.

The relation between the radius of impurity, radius of atom and body diagonal is shown as follows:

AD=2R+2r----(B)

   In    \Delta {\rm{ABC}},

(AB) ^2 =(AC) ^2 +(BC) ^2

For calculation of AD, AB is determined using Pythagoras theorem.

Substitute {\rm{AC}} as a and {\rm{BC}} as a in above equation as follows:

(AB) ^2 =a ^2 +a ^2

AB= \sqrt{2a} ----(1)

Also,

AB=2R

Substitute value of 2{\rm{R}} for {\rm{AB}} in equation (1) as follows:

2R= \sqrt{2} aa = \sqrt{2} R

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(AD) ^2 =(AB) ^2 +(BD)^2

Substitute {\rm{AB}} as \sqrt 2a   and {\rm{BD}} as a in above equation as follows:

(AD) ^2 =( \sqrt 2a) ^2 +(a) ^2 AD= \sqrt3a

For calculation of radius of an impure atom in FCC tetrahedral site,

Substitute value of AD in equation (B) as follows:

\sqrt 3a=2R+2r

Substitute a as \sqrt 2{\rm{R}} in above equation as follows:

( \sqrt3 )( \sqrt2 )R=2R+2r\\\\

r = \frac{2.4494R-2R}{2}\\

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Note

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