Question

If 10.0 liters of oxygen at stp are heated to 512 °c, what will be the new volume of gas if the pressure is also increased to 1520.0 mm of mercury?

Answer 1

Answer:

14.4 L

Explanation:

Initially, the gas is at stp (standard conditions), which means

$T_1 = 273 K\\p_1= 1.01 \cdot 10^5 Pa$

and its initial volume is

$V_1 = 10 L = 0.010 m^2$

Later, the gas is heated to a final temperature of

$T_2=512 C + 273 =785 K$

and the pressure is increased to

$p_2 = 1520.0 mmHg \cdot \frac{1.01\cdot 10^5 Pa}{760 mmHg}=2.02\cdot 10^5 Pa$

So we can use the ideal gas equation to find the new volume, V2:

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\\V_2 = \frac{p_1 V_1 T_2}{p_2 T_1}=\frac{(1.01\cdot 10^5 Pa)(0.010 L)(785 K)}{(2.02\cdot 10^5 Pa)(273 K)}=0.0144 m^2 = 14.4 L$