Question

Question #25bda

Answer 1

(c) `257` `"kg"`

Explanation:

We're asked to find the mass of `"O"_2` in a `1000` `"L"` tank at `27` `""^"o""C"` with an internal pressure of `2.0 xx 10^7` `"Pa"`.

Every time you're given three of the four basic characteristics of a gas (pressure, volume, temperature, and/or moles), you'll be using the ideal-gas equation:

`PV = nRT`

where

• `P` is the pressure of the gas, expressed in atmospheres (`"atm"`).

Since the pressure in this case is given in pascals, we have to convert this to atmospheres, using the conversion factor

`1` `"atm" = 101,325` `"Pa"`

`2.0 xx 10^7` `cancel("Pa")((1"atm")/(101,325cancel("Pa"))) = color(red)(197` `color(red)("atm"`

• `V` is the volume occupied by the gas, expressed in liters, which is given as `color(orange)(1000)` `color(orange)("L")`

• `n` is the quantity of gas present, in moles, which is what we must find in order to calculate the mass of oxygen present

• `R` is called the universal gas constant, and is equal to `color(purple)(0.08206 ("L · atm")/("mol · K")`

• `T` is the absolute temperature of the system, "absolute" indicating that the temperature is in units of Kelvin, `"K"`

Since our given temperature is in degrees Celsius, we have to convert this to Kelvin using the formula

`"K" = ""^"o""C" + 273`

`"K" = 27^"o""C" + 273 = color(green)(300)` `color(green)("K")`

Now that we have all our necessary units, let's use the ideal-gas equation to find the number of moles of `"O"_2` present, by rearranging the equation to solve for `n`:

`PV = nRT`

`n = (PV)/(RT)`

`n = ((197cancel("atm"))(1000cancel("L")))/((0.08206 (cancel("L") "·" cancel("atm"))/("mol" "·" cancel("K")))(300cancel("K"))) = 8018` `"mol O"_2`

Now that we know the moles of gas present, we can use the molar mass of `"O"_2`, `32.00"g"/"mol"`, to calculate the number of grams of oxygen:

`8018` `cancel("mol O"_2)((32.00"g O"_2)/(1cancel("mol O"_2))) = 2.566 xx 10^5` `"g O"_2`

Lastly, let's convert this to kilograms:

`2.566 xx 10^5` `cancel("g O"_2)((1"kg O"_2)/(10^3cancel("g O"_2))) = color(blue)(257` `color(blue)("kg O"_2`

Thus, the correct answer is (c).