# Question 25bda

Jun 4, 2017

(c) $257$ $\text{kg}$

#### Explanation:

We're asked to find the mass of ${\text{O}}_{2}$ in a $1000$ $\text{L}$ tank at $27$ $\text{^"o""C}$ with an internal pressure of $2.0 \times {10}^{7}$ $\text{Pa}$.

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:

$P V = n R T$

where

• $P$ is the pressure of the gas, expressed in atmospheres ($\text{atm}$).

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

$1$ $\text{atm} = 101 , 325$ $\text{Pa}$

$2.0 \times {10}^{7}$ 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 $\textcolor{\mathmr{and} a n \ge}{1000}$ $\textcolor{\mathmr{and} a n \ge}{\text{L}}$

• $n$ 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, $\text{K}$

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

$\text{K" = ""^"o""C} + 273$

$\text{K" = 27^"o""C} + 273 = \textcolor{g r e e n}{300}$ $\textcolor{g r e e n}{\text{K}}$

Now that we have all our necessary units, let's use the ideal-gas equation to find the number of moles of ${\text{O}}_{2}$ present, by rearranging the equation to solve for $n$:

$P V = n R T$

$n = \frac{P V}{R T}$

$n = \left(\left(197 \cancel{\text{atm"))(1000cancel("L")))/((0.08206 (cancel("L") "·" cancel("atm"))/("mol" "·" cancel("K")))(300cancel("K}}\right)\right) = 8018$ ${\text{mol O}}_{2}$

Now that we know the moles of gas present, we can use the molar mass of ${\text{O}}_{2}$, $32.00 \text{g"/"mol}$, 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 ${\text{g O}}_{2}$

Lastly, let's convert this to kilograms:

$2.566 \times {10}^{5}$ 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).