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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?

Respuesta :

skyluke89

Answer:

14.4 L

Explanation:

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

[tex]T_1 = 273 K\\p_1= 1.01 \cdot 10^5 Pa[/tex]

and its initial volume is

[tex]V_1 = 10 L = 0.010 m^2[/tex]

Later, the gas is heated to a final temperature of

[tex]T_2=512 C + 273 =785 K[/tex]

and the pressure is increased to

[tex]p_2 = 1520.0 mmHg \cdot \frac{1.01\cdot 10^5 Pa}{760 mmHg}=2.02\cdot 10^5 Pa[/tex]

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

[tex]\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[/tex]

dsdrajlin

If 10.0 liters of oxygen at STP are heated to 512 °C and its pressure increased to 1520.0 mmHg, the new volume will be 14.4 L.

10.0 L (V₁) of oxygen is at standard temperature (T₁ = 273.15 K) and standard pressure (P₁ = 760.0 mmHg).

It is heated to 512 °C (T₂) and the pressure increased to 1520.0 mmHg (P₂). We will convert 512 °C to Kelvin using the following expression.

[tex]K = \° C + 273.15 = 512\° C + 273.15 = 785 K[/tex]

Then, we can calculate the new volume (V₂) using the combined gas law.

[tex]\frac{P_1 \times V_1 }{T_1} = \frac{P_2 \times V_2 }{T_2} \\\\V_2 = \frac{P_1 \times V_1 \times T_2 }{T_1 \times P_2} = \frac{760.0 mmHg \times 10.0 L \times 785 K }{273.15 K \times 1520.0 mmHg} = 14.4 L[/tex]

If 10.0 liters of oxygen at STP are heated to 512 °C and its pressure increased to 1520.0 mmHg, the new volume will be 14.4 L.

Learn more: https://brainly.com/question/13154969

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