Question

For the algebra here, you can simply by working one side then isolating temperature 1 OR you can use algebra first - that part is up to you. Please answer first in unit of Kelvin temperature. Then subtract 273 to solve for the temperature in degrees Celsius. Units are provided, solve both answers to one decimal or the nearest tenths place. (1 point)A sample of a gas has its volume of 2.80 L at an unknown temperature when the sample is submerged in ice bath at 0.00 Celsius it's volume decreases to 2.7 what was it initial temperature in both Kevin and Celsius Next,

Answer 1

To solve for the initial temperature of the gas in Kelvin and Celsius, we can use Charles's Law. Charles's Law states that the volume of a gas is directly proportional to its temperature in Kelvin when pressure is held constant. The formula is given as:

$$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$

Where:

• $V_1$ = initial volume of the gas
• $T_1$ = initial temperature of the gas in Kelvin
• $V_2$ = final volume of the gas
• $T_2$ = final temperature of the gas in Kelvin

We know:

• $V_1=2.80,\text{L}$
• $V_2=2.70,\text{L}$
• $T_2=0.00,\text{C}=273.15,\text{K}$

We can rearrange the equation to solve for $T_1$:

$$T_1=\frac{V_1\times T_2}{V_2}$$

Now we can plug in the known values:

$$T_1=\frac{2.80,\text{L}\times 273.15,\text{K}}{2.70,\text{L}}$$

Calculating this gives:

$$T_1=\frac{764.82}{2.70}\approx 283.2,\text{K}$$

Next, converting $T_1$ to Celsius:

$$T_1(\text{C})=T_1(\text{K})-273.15$$

$$T_1(\text{C})\approx 283.2-273.15\approx 10.1,\text{C}$$

So, the initial temperature of the gas is approximately:

• Temperature in Kelvin: 283.2 K
• Temperature in Celsius: 10.1 °C