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baherus [9]
3 years ago
10

According to the combined gas law, the pressure exerted by a gas is:

Chemistry
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer : The correct option is, (d) inversely proportional to volume and directly proportional to the kelvin temperature

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Charles' Law : It is defined as the volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

Combined gas law : It is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{PV}{T}=K

or,

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

From the above information we conclude that the pressure exerted by a gas is inversely proportional to volume and directly proportional to the kelvin temperature.

Hence, the correct option is, (d)

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A 298 lb person must receive Heparin, determine the number of units given every hour (Heparin 8.0 units/kg per hour). Enter your
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A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
Assoli18 [71]

Answer:

The specific heat of copper is 0.385 J/g°C

Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

m(copper)*c(copper)*ΔT(copper) = - m(water) * c(water) * ΔT(water)

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

8 0
3 years ago
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