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

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

Step-by-step 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,

`(P_1V_1)/(T_1)=(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* V_1=2* 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:

`(0.99atm* 2.00L)/(273K)=(P_2* 4.00L)/(137K)`

`P_2=0.25atm`

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