Step 1: List the given values.
To convert the temperature from degree Celsius to kelvin, add 273 to the temperature expressed in degree Celsius.
[tex]\begin{aligned} V_1 & = \text{25.0 L} \\ T_1 & = 30^{\circ}\text{C} = \text{303 K} \\ V_2 & = \text{10.0 L} \end{aligned}[/tex]
Step 2: Calculate the final temperature (in degree Celsius) by using Charles' law.
To convert the temperature from kelvin to degree Celsius, subtract 273 from the temperature expressed in kelvin
[tex]\begin{aligned} \frac{V_1}{T_1} & = \frac{V_2}{T_2} \\ T_2V_1 & = T_1V_2 \\ \frac{T_2V_1}{V_1} & = \frac{T_1V_2}{V_1} \\ T_2 & = \frac{T_1V_2}{V_1} \\ & = \frac{(\text{330 K})(\text{10.0 L})}{\text{25.0 L}} \\ & = \text{121 K} \\ & = \boxed{-152^{\circ}\text{C}} \end{aligned}[/tex]
Hence, the gas will occupy 10.0 liters at -152°C.
[tex]\\[/tex]
#CarryOnLearning
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Verified answer
SOLUTION:
Step 1: List the given values.
To convert the temperature from degree Celsius to kelvin, add 273 to the temperature expressed in degree Celsius.
[tex]\begin{aligned} V_1 & = \text{25.0 L} \\ T_1 & = 30^{\circ}\text{C} = \text{303 K} \\ V_2 & = \text{10.0 L} \end{aligned}[/tex]
Step 2: Calculate the final temperature (in degree Celsius) by using Charles' law.
To convert the temperature from kelvin to degree Celsius, subtract 273 from the temperature expressed in kelvin
[tex]\begin{aligned} \frac{V_1}{T_1} & = \frac{V_2}{T_2} \\ T_2V_1 & = T_1V_2 \\ \frac{T_2V_1}{V_1} & = \frac{T_1V_2}{V_1} \\ T_2 & = \frac{T_1V_2}{V_1} \\ & = \frac{(\text{330 K})(\text{10.0 L})}{\text{25.0 L}} \\ & = \text{121 K} \\ & = \boxed{-152^{\circ}\text{C}} \end{aligned}[/tex]
Hence, the gas will occupy 10.0 liters at -152°C.
[tex]\\[/tex]
#CarryOnLearning