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Question 6
1 pts
In a system the pressure and mass are kept constant. The initial temperature of the system is \( -22.7^{\circ} \mathrm{C} \) and the volume is \( 18.4 \mathrm{~L} \). If the volume changes to \( 5.8 \mathrm{~L} \) what is the temperature in \( { }^{\circ} \mathrm{C} \) ?

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Answer

\( T_2 = \frac{V_2}{V_1} \cdot T_1 = \frac{5.8}{18.4} \cdot 250.45 = 78.903 \mathrm{K} \)

Steps

Step 1 ：\(V_1 = 18.4 \mathrm{~L}, \, T_1 = -22.7 + 273.15 = 250.45 \mathrm{K} \)

Step 2 ：\(V_2 = 5.8 \mathrm{~L}\)

Step 3 ：\( T_2 = \frac{V_2}{V_1} \cdot T_1 = \frac{5.8}{18.4} \cdot 250.45 = 78.903 \mathrm{K} \)