calculate the volume occupied by 32.0 g of o₂ gas, the pressure of the o₂ gas is 78.5 kpa at 25…

calculate the volume occupied by 32.0 g of o₂ gas, the pressure of the o₂ gas is 78.5 kpa at 25 °c.\nl\nround your answer to three significant figures. do not include units in your answer.\nenter the answer

calculate the volume occupied by 32.0 g of o₂ gas, the pressure of the o₂ gas is 78.5 kpa at 25 °c.\nl\nround your answer to three significant figures. do not include units in your answer.\nenter the answer

Answer

Explanation:

Step1: Convert masa a moles

Primero, calculamos los moles de $O_2$. La masa molar de $O_2$ es $M = 32.0\ g/mol$. Los moles $n$ se calculan como $n=\frac{m}{M}$, donde $m = 32.0\ g$. Entonces $n=\frac{32.0\ g}{32.0\ g/mol}=1\ mol$.

Step2: Convert temperatura a Kelvin

Convertimos la temperatura de Celsius a Kelvin. Usamos la fórmula $T(K)=T(^{\circ}C)+ 273.15$. Así, $T = 25^{\circ}C+273.15 = 298.15\ K$.

Step3: Convert presión a atmósferas

Convertimos la presión de $kPa$ a atmósferas. Sabemos que $1\ atm = 101.325\ kPa$. Entonces $P=\frac{78.5\ kPa}{101.325\ kPa/atm}\approx0.775\ atm$.

Step4: Aplicar ley de los gases ideales

La ley de los gases ideales es $PV = nRT$, donde $R = 0.0821\ L\cdot atm/(mol\cdot K)$. Despejamos $V$: $V=\frac{nRT}{P}$. Sustituyendo $n = 1\ mol$, $R=0.0821\ L\cdot atm/(mol\cdot K)$, $T = 298.15\ K$ y $P = 0.775\ atm$, tenemos $V=\frac{1\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times298.15\ K}{0.775\ atm}\approx31.7\ L$.

Answer:

31.7