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inysia [295]
3 years ago
13

A sample of helium gas at 27.0 °C and 3.60 atm pressure is cooled in the same container to a temperature of -73.0 °C. What is th

e new pressure, if volume and amount of gas do not change?
Chemistry
1 answer:
trasher [3.6K]3 years ago
5 0

Answer: The new pressure, if volume and amount of gas do not change is 2.40 atm

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=3.60atm\\T_1=27^0C=(27+273)K=300K\\P_2=?\\T_2=-73.0^0C=(273-73)=200K

Putting values in above equation, we get:

\frac{3.60}{300K}=\frac{P_2}{200K}\\\\P_2=2.40atm

Thus the new pressure, if volume and amount of gas do not change is  2.40 atm

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if a gas sample has a pressure of 761 mmhg at 0.00°c by how much does the temperature have to decrease to lower the pressure to
Nady [450]
<h3>Answer:</h3>

272.43 K or -0.718°C

<h3>Explanation:</h3>

We are given;

The initial pressure,P1 as 761 mmHg

Initial temperature, T1 as 0.00°C which is equivalent to 273.15 K

Final pressure as 759 mmHg

We are required to calculate the final temperature;

According to pressure law, the pressure of a gas and absolute temperature are directly proportional at constant volume.

That is; Pα T

Therefore, at varying pressure and temperature,

\frac{P1}{T1}=\frac{P2}{T2}

To get final temperature;

T2=\frac{P2T1}{P1}

T2=\frac{(759 mmHg)(273.15K)}{761 mmHg}

T2=272.43K

Therefore, the final temperature will be 272.43 K or -0.718°C

6 0
3 years ago
Question 11 please HELP
Lena [83]

The answer is A or B, I would put B.

6 0
3 years ago
Can someone please help me
Vanyuwa [196]

Answer:

1.7

Explanation:

Density = M/V

When you divide 4.52 by 2.6, you get 1.738461538, which can be simplified to 1.7.

7 0
3 years ago
108.95 g/mol, C3H2Cl2 Express your answer as a chemical formula.​
fredd [130]

Answer:

gchxlhxhkxhkxohxhochlc

6 0
3 years ago
A 400. 0 g sample of liquid water is at 30. 0 ºc. how many joules of energy are required to raise the temperature of the water t
Agata [3.3K]

Energy required to raise the temperature from 35°C - 45 °C= 25116 J.

specific heat, the quantity of warmth required to raise the temperature of one gram of a substance by means of one Celsius degree. The units of precise warmth are generally energy or joules consistent with gram according to Celsius diploma. for instance, the unique warmth of water is 1 calorie (or 4.186 joules) according to gram in step with Celsius degree.

solving,

Sample of liquid = 400. 0 g

temperature = 30. 0 ºc

joules of energy are required to raise the temperature of the water to 45. 0 ºc

therefore rise in temperature 45 - 30 = 15°C

Specific heat capacity = 4.186 J/g m °C

In kelvin = 273 + 15 = 288

          = ∴ energy required = Q = m s ( t final - t initial)

            =  400*4.186 * 15

           = 25116 joule

Learn more about specific heat here:-brainly.com/question/21406849

#SPJ4

7 0
1 year ago
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