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The volume of a sample of oxygen is 195.8 mL when the pressure is 0.146 am and the temperature is 28.6 °C. At what temperature is the volume 768.4 L and the pressure 0.455 am?

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To find the temperature at which the volume is 768.4 L and the pressure is 0.455 atm, we can use the combined gas law. The combined gas law combines Boyle's law, Charles's law, and Gay-Lussac's law into one equation:

(P1 * V1) / T1 = (P2 * V2) / T2

where:

P1 = Initial pressure (in atm)

V1 = Initial volume (in liters)

T1 = Initial temperature (in Kelvin)

P2 = Final pressure (in atm)

V2 = Final volume (in liters)

T2 = Final temperature (in Kelvin)

Let's plug in the given values:

P1 = 0.146 atm

V1 = 195.8 mL = 0.1958 L (converted to liters)

T1 = 28.6 °C = 28.6 + 273.15 = 301.75 K (converted to Kelvin)

P2 = 0.455 atm

V2 = 768.4 L

T2 = ? (to be calculated)

Now, let's rearrange the equation to solve for T2:

T2 = (P1 * V1 * T2) / (P2 * V2)

Now, substitute the values:

T2 = (0.146 * 0.1958 * 301.75) / (0.455 * 768.4)

Calculate the value of T2:

T2 ≈ 49.13 °C

Therefore, at a temperature of approximately 49.13 °C, the volume of the oxygen sample will be 768.4 L, and the pressure will be 0.455 atm.

User Rajasaur
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