Question

Inicialmente una muestra de neón tiene un volumen de 2,50 L a 15 °C. ¿Cuál es la

temperatura en °C, cuando el volumen de la muestra cambia a cada uno de los
siguientes volúmenes, ¿manteniéndose la presión constante?
a. 5,0 L b. 1250 ml c. 7,50L d. 3550 ml

Answer 1

Using the formulas derived from Charles's law, we calculated the corresponding temperatures in Celsius for the volume changes of neon at constant pressure. By converting given temperatures to Kelvin, performing calculations, and then converting back to Celsius, we found the final temperatures for each volume change.

Step-by-step explanation:

The student's question is related to the effect of temperature on gas volume under constant pressure, according to Charles's law. In each scenario, we're asked to find the temperature when the volume changes while keeping the pressure constant.

We can use the following formula, derived from Charles's law, which relates the initial and final volumes (V1 and V2) to the initial and final temperatures (T1 and T2) in Kelvin:

V1/T1 = V2/T2

This can be rearranged to find the final temperature:

T2 = (V2/V1) * T1

To find the temperature in Celsius, we subtract 273.15 from the Kelvin temperature. We need to convert all temperatures to Kelvin by adding 273.15 to the Celsius temperatures given.

Calculation for each volume:

1. Initial temperature in Kelvin: 15°C + 273.15 = 288.15 K
2. 5.0 L: T2 = (5.0 L / 2.50 L) * 288.15 K = 576.3 K, so in °C it's 576.3 K - 273.15 = 303.15°C
3. 1250 ml (1.25 L): T2 = (1.25 L / 2.50 L) * 288.15 K = 144.075 K, so in °C it's 144.075 K - 273.15 = -129.075°C
4. 7.50 L: T2 = (7.50 L / 2.50 L) * 288.15 K = 864.45 K, so in °C it's 864.45 K - 273.15 = 591.3°C
5. 3550 ml (3.55 L): T2 = (3.55 L / 2.50 L) * 288.15 K = 414.258 K, so in °C it's 414.258 K - 273.15 = 141.108°C

Answer 2

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