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alexandr1967 [171]
3 years ago
5

A sample of gas in a sealed container (fixed volume) is heated from room temperature to 80.0°C.

Chemistry
1 answer:
inna [77]3 years ago
8 0
Hello!

A) When the temperature inside the container increases, the pressure inside the container increases as well. 

An example for that is when you have an aerosol can and start heating it. The pressure of the gas inside the aerosol can will start to increase, and that would lead to the exploding of the can if heating is kept for too long. Bombs work on this principle too: The heat from the violent chemical reaction inside the closed compartment increase the pressure of the gases until the fragments are ejected at high velocities. 

B) The effect of temperature on the pressure of a gas illustrates Gay-Lussac's Law.

This law was formulated by the famous French chemist that gives it its name. It relates the expansion of a gas with the increase in temperature when the volume is left constant. The Gay-Lussac's Law can be expressed as follows, for the case of this exercise:

\frac{P1}{T1}= \frac{P2}{T2} \\ \\ P2= \frac{P1}{T1}*T2=P1* \frac{80 degC}{25degC}= P1*3,2

You can see that the factor that is multiplying P1 is higher than 1 for the case of heating from 25 °C to 80 °C, so the pressure will increase.

C) At a molecular level, when the temperature is raised the kinetic energy of the molecules inside the container will increase. This increase in the kinetic energy will cause the molecules to move faster, and to hit the walls of the container more often. This causes an increase in the pressure inside the container because there more hits means more force on the walls of the container, and that is the definition of pressure. 

Have a nice day!
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n = 0.9924 moles of NaCl

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Therefore:

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Molarity = 0.9924 M

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Answer:

The new partial pressures after equilibrium is reestablished:

PCl_3,p_1'=6.798 Torr

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Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

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