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Chapter 12: Problem 41

Use Charles's law to solve for the missing value in the following. \(V_{1}=80.0 \mathrm{mL}, T_{1}=\) \(27^{\circ} \mathrm{C}, T_{2}=77^{\circ} \mathrm{C}, V_{2}=?\)

Short Answer

Expert verified
The value of \(V_2\), which is the volume at the new temperature, is calculated to be approximately 93.39 mL.

Step by step solution

01

Convert temperatures from Celsius to Kelvin

Before applying Charles's Law, it is crucial to convert the temperatures from degrees Celsius to Kelvin. The conversion is done by the formula: K = C + 273.15. Hence, \(T_1 = 27\degree C = 27 + 273.15 = 300.15 K\), and \(T_2 = 77\degree C = 77 + 273.15 = 350.15 K\)
02

Substitute into Charles's Law

Now that temperatures are in absolute scale, Charles' Law can be applied:\((80.0 mL) / (300.15 K) = V_2 / (350.15 K)\). Cross-multiply to solve for the unknown variable \(V_2 = \((80.0 mL) \times (350.15 K)) / (300.15 K).\)
03

Calculate Missing Value

Perform known algebraic calculations to get the value of \(V_2\).

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Most popular questions from this chapter

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