The metal lid on a plastic container is
sealed at 2.3 atm of pressure at a
temperature of 298K. The container
is placed in the trunk of a car on a
cold, frigid day (T= 2°C). What is
the pressure of the gas inside the can
at this temperature?

2.12 atm

Explanation:

Given data:

Initial pressure = 2.3 atm

Initial temperature = 298 K

Final temperature = 2°C = 2+273 = 275 k

Final pressure = ?

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

2.3 atm / 298 K = P₂/275 K

P₂ = 2.3 atm × 275 K / 298 K

P₂ = 632.5  atm. K /298 K

P₂ = 2.12 atm

The pressure at 2°C is 2.12 atm.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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