Question

A sample of argon gas has a volume of 735 mL at a pressure of 1.20 atm and a temperature of 112 ∘C. What is the final volume of the gas, in milliliters, when the pressure and temperature of the gas sample are changed to the following, if the amount of gas does not change? 0.59 atm and 68 ∘C

Answer 1

`1,324\text{ mL}`

Step-by-step explanation:

Here, we want to get the final volume

Mathematically, from the general gas formula:

`(P_1V_1)/(T_1)\text{ = }(P_2V_2)/(T_2)`

P1 is the initial pressure which is 1.2 atm

V1 is the initial volume which is 735 mL

T1 is the initial temperature which is 112 degrees Celsius (we convert to Kelvin by adding 273: 273 + 112 = 385 k)

P2 is the final pressure which is 0.59 atm

V2 is the final volume which is unknown

T2 is the final temperature which is 68 + 273 = 341 K

Substituting the values, we have it that:

`\begin{gathered} (1.2*735)/(385)\text{ = }(0.59* V_2)/(341) \\ \\ V_2\text{ = }(341*1.2*735)/(385*0.59)\text{ = 1324 mL} \end{gathered}`