2) A sample of argon has a volume of 5.0 Land the pressure is 650 mm Hg If the final temperature is
30. C, the final volume is 5.7 L, and the final pressure is 800. mm Hg, what was the initial temperature
of the argon?

Answers

Answer 1

Answer:

[tex]T_1 = -57.15^{\circ}C[/tex]

Explanation:

Given

[tex]P_1 = 650mmHg[/tex] --- Initial Pressure

[tex]V_1 = 5.0L[/tex] --- Initial Volume

[tex]V_2 = 5.7L[/tex] --- Final Volume

[tex]P_2 = 800mmHg[/tex] --- Final Pressure

[tex]T_2 = 30C[/tex] ---- Final Temperature

Required

Determine the initial temperature (T1)

This question will be solved using combined gas law which states:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

However, the final temperature must be converted to degree kelvin

[tex]T_2 = 30C[/tex] --- Add 273.15

[tex]T_2 = 30k + 273.15 k[/tex]

[tex]T_2 = 303.15k[/tex]

Make T1 the subject in [tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

[tex]T_1 = \frac{P_1V_1T_2}{P_2V_2}[/tex]

Substitute values for P1, V1, T2, P2 and V2

[tex]T_1 = \frac{650 * 5.0 * 303.15}{800 * 5.7}[/tex]

[tex]T_1 = \frac{985237.5}{4560}[/tex]

[tex]T_1 = 216.060855263[/tex]

Approximate

[tex]T_1 = 216k[/tex]

Convert to degree Celsius

[tex]T_1 = 216k[/tex] --- Subtract 273.15

[tex]T_1 = 216 - 273.15[/tex]

[tex]T_1 = -57.15^{\circ}C[/tex]

Hence, the initial temperature is -57.15C


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