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Aliun [14]
3 years ago
11

A gas originally at 27 °c and 1.00 atm pressure in a 2.6 l flask is cooled at constant pressure until the temperature is 11 °c.

the new volume of the gas is
Chemistry
1 answer:
elena55 [62]3 years ago
8 0
Charles law gives the relationship between volume and temperature of gas at constant pressure 
it states that at constant pressure, volume of gas is directly proportional to temperature 
V/T = k
where V - volume T - temperature and k - constant 
\frac{V1}{T1} =  \frac{V2}{T2}
parameters for the first instance are on the left side of the equation and parameters for the second instance are on the right side of the equation
T1 - temperature in Kelvin - 27 °C + 273 = 300 K
T2 - 11 °C + 273 = 284 K
substituting the values in the equation 
2.6 L / 300 K = V / 284 K 
V = 2.46 L
New volume of the gas is 2.46 L
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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
Goryan [66]

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}

Then, the pressure increases to 1.58 atmospheres and the volume decreases to 2.46 cubic meters.

\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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3 years ago
A 3.20-mol sample of gas occupies a volume of 350. mL at 300.0 K. Determine the
nalin [4]

Answer:

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Explanation:

From the question we are told that:

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 PV=nRT

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