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STALIN [3.7K]
3 years ago
10

In a separate experiment beginning from the same initial conditions, including a temperature Ti of 25.0°C, half the number of mo

les found in part (a) are withdrawn while the temperature is allowed to vary and the pressure undergoes the same change from 25.7 atm to 4.10 atm.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?
Chemistry
1 answer:
NNADVOKAT [17]3 years ago
6 0

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

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What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 3.0 ATM and the temperatu
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Answer:

\boxed {\boxed {\sf P_2= 9 \ atm}}

Explanation:

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\frac{450 \ atm*mL}{50 \ mL}= \frac{P_2 * 50 \ mL}{50 \ mL}

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The units of milliliters will cancel.

\frac{450 \ atm}{50 }= P_2

9 \ atm =P_2

The new pressure is <u>9 atmospheres.</u>

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