Question

In a separate experiment beginning from the same initial conditions, including a temperature Ti of 25.0°C, half the number of moles found in part (a) are withdrawn while the temperature is allowed to vary and the pressure undergoes the same change from 25.7 atm to 4.10 atm. What is the final temperature (in °C) of the gas?

Answer 1

Answer:

- 178 ºC

Explanation:

The  ideal gas law   states that :

PV = nRT,

where P is the pressure, V is the volume, n is number of moles , R is the gas constant and T is the absolute temperature.

For the initial conditions :

P₁ V₁ = n₁ R T₁    (1)

and for the final conditions:

P₂V₂= n₂ R T₂    where   n₂ = n₁/2     then    P₂ V₂ = n₁/2 T₂    (2)

Assuming V₂ = V₁ and  dividing (2) by Eqn (1) :

P₂ V₂ = n₁/2 R T₂  / ( n₁ R T₁)      then  P₂ / P₁ = 1/2 T₂ / T₁

4.10 atm / 25.7 atm = 1/2 T₂ / 298 K ⇒ T₂ = 0.16 x 298 x 2 = 95.1 K

T₂ = 95 - 273 = - 178 º C

Answer 2

The final temperature (in °C) of the gas should be considered as the - 178 ºC.

Calculation of the final temperature:

Here we applied ideal gas law:

Since

PV = nRT,

here P is the pressure,

V is the volume, n is number of moles ,

R is the gas constant

and T is the absolute temperature.

Now

For the initial conditions :

P₁ V₁ = n₁ R T₁    (1)

and for the final conditions:

P₂V₂= n₂ R T₂    

where   n₂ = n₁/2    

So,     P₂ V₂ = n₁/2 T₂    (2)

Here we presume

V₂ = V₁ and  dividing (2) by Eqn (1) :

P₂ V₂ = n₁/2 R T₂  / ( n₁ R T₁)      after this  P₂ / P₁ = 1/2 T₂ / T₁

4.10 atm / 25.7 atm = 1/2 T₂ / 298 K ⇒ T₂ = 0.16 x 298 x 2 = 95.1 K

So,

T₂ = 95 - 273 = - 178 º C

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