Question

a glass container was initially charged with 1.50 mol of a gas sample at 3.75 atm and 21.7C. some of the gas was release as the temp was increased to 28.1C so the final pressure in the container was reduced to 0.998 atm. How many moles of the gas sample are present at the end?

Answer 1

Answer:

0.39 mol

Explanation:

Considering the ideal gas equation as:

$PV=nRT$

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

At same volume, for two situations, the above equation can be written as:-

$\frac {{n_1}\times {T_1}}{P_1}=\frac {{n_2}\times {T_2}}{P_2}$

Given ,  

n₁ = 1.50 mol

n₂ = ?

P₁ = 3.75 atm

P₂ = 0.998 atm

T₁ = 21.7  ºC

T₂ = 28.1 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (21.7 + 273.15) K = 294.85 K  

T₂ = (28.1 + 273.15) K = 301.25 K  

Using above equation as:

$\frac{{n_1}\times {T_1}}{P_1}=\frac{{n_2}\times {T_2}}{P_2}$

$\frac{{1.50\ mol}\times {294.85\ K }}{3.75\ atm}=\frac{{n_2}\times {301.25\ K  }}{0.998\ atm}$

$n_2=\frac{{1.50}\times {294.85}\times 0.998}{3.75\times 301.25}\ mol$

Solving for n₂ , we get:

n₂ = 0.39 mol

Answer 2

Taking into account the ideal gas law, 0.39 moles of the gas sample are present at the end.

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Then, the volume can be expressed as:

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

You have the same volume for both situations because the gas container did not change, so for the initial situation 1 and final 2 it can be expressed:

V1=V2

$\frac{n1xRxT1}{P1}=\frac{n2xRxT2}{P2}$

Being the value of R a constant, this last expression can be written as:

$\frac{n1xT1}{P1}=\frac{n2xT2}{P2}$

In this case, you know:

• n1= 1.50 moles
• T1= 21.7 C= 294.7 K
• P1=3.75 atm
• n2= ?
• T2= 28.1 C= 301.1 K
• P2= 0.998 atm

Replacing:

$\frac{1.50 molesx294.7 K}{3.75 atm}=\frac{n2x301.1 K}{0.998 atm}$

Solving:

$n2=\frac{0.998 atm}{301.1 K} x\frac{1.50 molesx294.7 K}{3.75 atm}$

n2= 0.39 moles

Finally, 0.39 moles of the gas sample are present at the end.

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