Question

500.0 L of gas are prepared at 0.921 atm pressure and 200.0°C. The gas is placed into a tank under high pressure. When the tank cools to 20.0°C, the pressure is 30.0 atm. What is the volume of the gas under these conditions?

Answer 1

`"V" = 24.8 color(white)(l) "L"`

Explanation:

The ideal gas law suggests the relationship

`P * V = n * R * T`

that relates

• `P`, the pressure of the gas
• `V`, the volume the gas occupies
• `n`, the number of moles of gas in the closed container
• `T`, the temperature of the gas in `color(purple)("degrees Kelvin")`, and
• `R` the ideal gas constant.

Let `n_1`, `P_1`, `V_1`, and `T_1` resembles the respective properties of the gas before the change and `n_2`, `P_2`, `V_2`, and `T_2` properties after the change.

• `n_1 = (R * T_1) / (P_1 * V_1)`

• `n_2 = (R * T_2) / (P_2 * V_2)`

The amount of gas present in the container, `n`, stays the same during this process. `n_1 = n_2` and therefore

`(R * T_1) / (P_1 * V_1) = n = (R * T_2) / (P_2 * V_2)`

`(T_1) / (P_1 * V_1) = (T_2) / (P_2 * V_2)`

Multiplying both sides of the equation by `(P_1 * V_2) / (T_1)` gives the relationship between the volume of the gas before and after the change:

`(V_2) / (V_1) = (T_2) / (T_1) * (P_1) / (P_2)`

Therefore

`V_2 = V_1 * ((T_2) / (T_1) * (P_1) / (P_2))`
`color(white)(V_2) = 500.0 color(white)(l) "L" xx ((200 + 273.15) color(white)(l) color(red)(cancel(color(purple)(K)))) / ((20 + 273.15) color(white)(l) color(red)(cancel(color(purple)(K)))) * (0.921 color(white)(l) color(red)(cancel(color(black)("atm")))) / (30.0 color(white)(l) color(red)(cancel(color(black)("atm"))))`
`color(white)(V_2) = 24.8 color(white)("L")`