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