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Zorluk: OrtaGas Laws and the Ideal Gas Equation

A diver releases a bubble of air of volume 8.00 cm38.00\text{ cm}^3 at a depth where the water pressure is 3.50×105 Pa3.50 \times 10^5\text{ Pa} and the temperature is 7C7^\circ\text{C}. What is the volume of the air bubble, in cm3\text{cm}^3, just as it reaches the surface where the pressure is 1.00×105 Pa1.00 \times 10^5\text{ Pa} and the temperature is 27C27^\circ\text{C}?

Cevap: 30 cm³

Cevap

The final volume of the air bubble at the surface is 30.0 cm330.0\text{ cm}^3.
Using the combined gas law P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} with absolute temperatures T1=280 KT_1 = 280\text{ K} and T2=300 KT_2 = 300\text{ K} yields a final volume of 30.0 cm330.0\text{ cm}^3.

Adım Adım Çözüm

1
Convert temperatures from Celsius to Kelvin
T1=7C+273=280 KT_1 = 7^\circ\text{C} + 273 = 280\text{ K} and T2=27C+273=300 KT_2 = 27^\circ\text{C} + 273 = 300\text{ K}
Gas laws require absolute thermodynamic temperature in Kelvin.
2
Apply the combined gas law equation
P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
The amount of gas in the bubble remains constant while pressure, volume, and temperature all change simultaneously.
3
Rearrange for the unknown volume V2V_2 and substitute the values
V2=3.50×105×8.00×3001.00×105×280=30.0 cm3V_2 = \frac{3.50 \times 10^5 \times 8.00 \times 300}{1.00 \times 10^5 \times 280} = 30.0\text{ cm}^3
Calculates the expanded volume of the air bubble at surface conditions.

Anahtar Kavram

Combined Gas Law
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