Question

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

Answer 1

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:

    P₁V₁   =    P₂V₂

      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

Answer 2

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.