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ddd [48]
3 years ago
6

according to Boyles law if the initial pressure and volume of a gas were five atm for the pressure and 10 mL for the volume and

pressure is doubled what is the new volume
Chemistry
2 answers:
maks197457 [2]3 years ago
5 0

Answer:The new volume is 5mL

Explanation:

The formular for Boyles Law is; P1 V1 = P2 V2

Where P1 = 1st Pressure   V1 = First Volume

          P2 = 2nd Pressure V2 = Second Volume

From the question; P1 = 5atm, V1 = 10ml

                                P2 = 2 x P1 (2 x 5) = 10 atm   V2 =?  

Using the Boyles Law Formular;  P1 V1 = P2 V2, we make V2 the subject of formular;  P1 V1/ P2 = V2

∴ 5 x 10/ 10 = 5

∴ V2 = 5mL

Gre4nikov [31]3 years ago
4 0

Answer:

The initial volume has been halved to 5 mL

Explanation:

Step 1: Data given

Initial pressure = 5.0 atm

Initial volume = 10 mL = 0.010 L

The pressure is doubled to 10.0 atm

Step 2: Calculate the new volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure of the gas = 5.0 atm

⇒with V1 = the initial volume of the gas = 10 mL = 0.010 L

⇒with P2 = the new pressure = 10.0 atm

⇒with V2 = the new volume = TO BE DETERMINED

5.0 atm * 0.010 L = 10.0 atm * V2

V2 = 0.005 L = 5 mL

Initial volume / final volume = 10 mL / 5 mL  = 2

The initial volume has been halved to 5 mL

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Define the term inertia
n200080 [17]

Answer:

Explanation:

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5 0
3 years ago
Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0
3 years ago
A 2.5 L container is filled with propane. The ambient temperature is 25°C and the
madam [21]

Answer:

The pressure inside the container will be 3.3 atmospheres

Explanation:

The relationship between the temperature and pressure of a gas occupying a fixed volume is given by Gay-Lussac's law which states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is kept constant.

Mathematically, it expressed as: P₁/T₁ = P₂/T₂

where P₁ is initial pressure, T₁ is initial temperature, P₂ is final pressure, T₂ is final temperature.

The above expression shows that the ratio of the pressure and temperature is always constant.

In the given question, the gas in the can attains the temperature of its environment.

P₁ = 3 atm,

T₁ = 25 °C = (273.15 + 25) K = 298.15 K,

P₂ = ?

T₂ = (55 °C = 273.15 + 55) K = 328.15 K

Substituting the values in the equation

3/298.15 = P₂/328.15

P₂ = 3 × 328.15/298.15

P₂ = 3.3 atm

Therefore, the pressure inside the container will be 3.3 atmospheres

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