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A gas has an initial volume of 3,480 no and an initial temperature of –70.0°C. What must be the temperature of the gas in Kelvin if its volume is reduced to 2,450 ml?
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Sagot :

SOLUTION:

Step 1: List the given values.

To convert the temperature from degree Celsius to kelvin, add 273.15 to the temperature expressed in degree Celsius.

[tex]\begin{aligned} V_1 & = \text{3,480 mL} \\ T_1 & = -70.0^{\circ}\text{C} = \text{203.15 K} \\ V_2 & = \text{2,450 mL} \end{aligned}[/tex]

Step 2: Calculate the final temperature (in kelvin) by using Charles' law.

[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{203.15 K})(\text{2,450 mL})}{\text{3,480 mL}} \\ & = \boxed{\text{143.0 K}} \end{aligned}[/tex]

Hence, the temperature of the gas is 143.0 K.

[tex]\\[/tex]

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