Question #3d499

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Question #3d499

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Answer 1 · 2017-05-02T01:40:08

$T_{2} = 381^{\circ} K$ $T_2 = 381^@"K"$

Explanation:

Convert the temperature to Kelvin scale:

$T_{1} = 32^{\circ} C + 273^{\circ} C = 305^{\circ} K$ $T_1 = 32^@"C"+273^@"C" = 305^@"K"$

Start with the combined Gas Law:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1 = (P_2V_2)/T_2$

Solve for $T_{2}$ $T_2$

$T_{2} = \frac{T_{1} P_{2} V_{2}}{P_{1} V_{1}}$ $T_2 = (T_1P_2V_2)/(P_1V_1)$

$T_{2} = \frac{(305^{\circ} K) (1.45 \times 10^{3} mm Hg) (108 ml)}{(745 mm Hg) (168 ml)}$ $T_2 = ((305^@"K")(1.45xx10^3"mm Hg")(108"ml"))/((745"mm Hg")(168"ml"))$

$T_{2} = 381^{\circ} K$ $T_2 = 381^@"K"$

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Question #3d499

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