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VMariaS [17]
3 years ago
10

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

evation where the pressure is 0.75 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?
Chemistry
2 answers:
gavmur [86]3 years ago
6 0

Answer:

The temperature of the air at this elevation is 335.2 K or 62.2 °C

Explanation:

Step 1: Data given

The initial volume = 1.2 L

The initial temperature = 25.0 °C = 298 K

The pressure = 1.0 atm

The pressure reduces to 0.75 atm

The volume increases to 1.8 L

Step 2: Calculate the new temperature

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

⇒with P1 = the initial pressure in the balloon = 1.0 atm

⇒with V1 = the initial volume of the balloon = 1.2 L

⇒with T1 = the initial temperature = 298 K

⇒with P2 = the reduced pressure = 0.75 atm

⇒with V2 = the increased volume = 1.8 L

⇒with T2 = the new volume = TO BE DETERMINED

(1.0 atm * 1.2L) / 298K = (0.75 atm * 1.8 L) / T2

0.004027 = 1.35 / T2

T2 = 1.35 / 0.004027

T2 = 335.2 K

335.2 - 273 = 62.2 °C

The temperature of the air at this elevation is 335.2 K = 62.2 °C

musickatia [10]3 years ago
4 0

Answer:

T_2=335.42K=62.27^oC

Explanation:

Hello,

In this case, by using the general gas law, that allows us to understand the pressure-volume-temperature relationship as shown below:

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

Thus, solving for the temperature at the end (considering absolute units of Kelvin), we obtain:

T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{1.8L*0.75atm*(25+273.15)K}{1.2L*1.0atm} \\\\T_2=335.42K=62.27^oC

Best regards.

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<h3>Answer:</h3>

        NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄  =  132.06 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age  =  21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age  =  21.18 %

In KNO₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of KNO₃  =  101.10 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of KNO₃ × 100

Mass %age  =  14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age  =  13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄  =  115.03 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age  =  14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age  =  12.17 %

In NH₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of NH₃  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₃ × 100

Mass %age  =  14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age  =  82.20 %

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Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of NH₄NO₃  =  80.04 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age  =  28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

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