2. if 10.0 liters of oxygen at stp are heated to 512 °c, what will be the new volume of gas if the pressure…

Answer:

We need to use the combined - gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$. At STP, $P_1 = 760\ mmHg$, $T_1=273\ K$, $V_1 = 10.0\ L$, $P_2=1520\ mmHg$, $T_2=(512 + 273)\ K=785\ K$.

First, substitute the values into the combined - gas law formula: [V_2=\frac{P_1V_1T_2}{P_2T_1}]

Step1: Identify the values

$P_1 = 760\ mmHg$, $V_1 = 10.0\ L$, $T_1 = 273\ K$, $P_2 = 1520\ mmHg$, $T_2=785\ K$

Step2: Substitute into the formula

[V_2=\frac{760\times10.0\times785}{1520\times273}]

[V_2=\frac{7600\times785}{415960}]

[V_2=\frac{5966000}{415960}\approx14.34\ L]

Explanation:

Step1: List given values

List pressure, volume and temperature values.

Step2: Apply combined - gas law

Use $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ to find $V_2$.