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netineya [11]
3 years ago
5

Un buzo que nada en las profundidades despide pequeñas burbujas de aire¿Aumenta disminuye o permanece igual el tamaño de las bur

bujas conforme suben a la superficie?
Chemistry
1 answer:
alekssr [168]3 years ago
5 0

Answer:

Al ascender las burbujas van aumentando de tamaño.

Explanation:

Las burbujas que produce el buzo debajo del agua son pequeñas moléculas de dióxido de carbono gaseoso producto de la respiración del mismo.

Ahora, a medida que las burbujas suben a la superficie, la presión que sufren estas (Presión debido al agua), es menor conforme van ascendiendo debido a la ley de Boyle: A medida que la presión aumenta, el volumen va disminuyendo.

Esto significa que al ascender las burbujas van aumentando de tamaño debido a que la presión que sufren estas es menor que cuando están a mayores profundidades.

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A gas has a volume of 4.25 m3 at a temperature of 95.0°C and a pressure of 1.05 atm. What temperature will the gas have at a pre
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Answer:

\boxed {\boxed {\sf 82.7 \textdegree C}}

Explanation:

We are asked to find the temperature of a gas given a change in pressure and volume. We will use the Combined Gas Law, which combines 3 gas laws: Boyle's, Charles's, and Gay-Lussac's.

\frac {P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Initially, the gas has a pressure of 1.05 atmospheres, a volume of 4.25 cubic meters, and a temperature of 95.0 degrees Celsius.

\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{P_2V_2}{T_2}

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\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{1.58  \ atm *2.46 \ m^3}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator by the second denominator, then multiply the first denominator by the second numerator.

(1.05 \ atm * 4.25 \ m^3) * T_2 = (95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)

Now the variable is being multiplied by (1.05 atm * 4.25 m³). The inverse operation of multiplication is division, so we divide both sides by this value.

\frac {(1.05 \ atm * 4.25 \ m^3) * T_2}{(1.05 \ atm * 4.25 \ m^3)} = \frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}

T_2=\frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}

The units of atmospheres and cubic meters cancel.

T_2=\frac{(95.0 \textdegree C)*(1.58* 2.46 )}{(1.05 * 4.25 )}

Solve inside the parentheses.

T_2= \frac{(95.0 \textdegree C)*3.8868}{4.4625}

T_2= \frac{369.246}{4.4625} \textdegree C}

T_2 = 82.74420168 \textdegree C

The original values of volume, temperature, and pressure all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 4 in the hundredth place to the right tells us to leave the 7 in the tenths place.

T_2 \approx 82.7 \textdegree C

The temperature is approximately <u>82.7 degrees Celsius.</u>

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