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kakasveta [241]
3 years ago
13

Some oxygen gas takes up a volume of 2.00 liters at 0.99 atm and 273 K. Its volume doubles and its temperature decreases to 137

K. What is the final pressure of the gas?
A.cannot be ddtermined
B.0.25 atm
C.0.98 atm
D.3.9 atm
Chemistry
2 answers:
nataly862011 [7]3 years ago
5 0

Answer:

The answer to your question is : letter B. 0.25 atm

Explanation:

To solve this problem we need to use the combined gas law:

    <u>P₁V₁</u>   =    <u>P₂V₂</u>

      T₁              T₂

Data

P1 = 0.99 atm  V1 = 2 l  T1 = 273K

P2 = ?               V2 = 4 l   T2 = 137K

Now, the clear P2 from the equation and we get

       P2 = P1V1T2 / T1V2

Substitution         P2 = (2 x 0.99 x 137)/(273 x 4)

       P2 = 271.26 / 1092

Result       P2 = 0.248 atm ≈ 0.25 atm

fredd [130]3 years ago
5 0

Answer : The correct option is, (B) 0.25 atm

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.99 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 2.00 L

V_2 = final volume of gas = 2\times V_1=2\times 2.00L=4.00L

T_1 = initial temperature of gas = 273 K

T_2 = final temperature of gas = 137 K

Now put all the given values in the above equation, we get:

\frac{0.99atm\times 2.00L}{273K}=\frac{P_2\times 4.00L}{137K}

P_2=0.25atm

Therefore, the final pressure of the gas is 0.25 atm.

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