Skip to player
Skip to main content
Search
Connect
Watch fullscreen
Like
Bookmark
Share
More
Add to Playlist
Report
Calculate the root mean square speed of hydrogen molecules at 0 degree centigrade and 1 atm pressure. Assuming hydrogen to be an ideal gas. The density of hydrogen is 8.99×10^-2 kg/m3
Physics with Akhtiar Ali Bullo
Follow
13 minutes ago
Physics class 12 numericals
12 class physics problems
Molecular Theory of Gases Numericals
Category
📚
Learning
Transcript
Display full video transcript
00:00
Numerical number 8 of unit number 15 molecular theory of gases is. Calculate the root mean
00:16
square speed of hydrogen molecules at 0 degree centigrade and 1 atm pressure. Assuming hydrogen
00:26
to be an ideal guess, the density of hydrogen is 8.99 into 10 raised power minus 2 kilogram
00:36
per meter cube. So first of all we will make a data,
00:43
the temperature obtained is 0 degree centigrade, 0 degree centigrade is also equal to 273 Kelvin
01:11
and pressure obtained is 1 atm. We can also write 1 atm as 1.01 into 10 raised power 5
01:30
Pascal. Our density of hydrogen is given is means rho is given is 8.99 into 10 raised power
01:48
minus 2 kilogram per meter cube is, mass of hydrogen molecule is equal to 2 into 1.67 into 10 raised power
02:14
minus 27 kilogram. So, mass of hydrogen molecule will be equal to 3.34 into 10 raised power minus
02:32
27 kilogram. And ball's main constant k is equal to 1.38 into 10 raised power minus 23
02:50
joule per Kelvin. Hameh kya find karna hai? Hameh root mean square speed find karna hai. So,
02:57
root mean square speed will be equal to watt. So, for calculation,
03:10
we will use two methods, means we can make this calculation by two methods. So, first method
03:27
root mean square speed is calculated by using formula under root 3 kt upon m. So, by putting the given values,
03:48
1.38 into 10 raised power minus 23 joule per Kelvin into temperature which is 273 Kelvin divided by mass of hydrogen
04:14
which is 3.34 into 10 raised power minus 27 kilogram. So, root mean square speed will be equal to
04:32
1.38 into 10 raised power. So, we will use 1.38 into 10 raised power minus 23 divided by 3.34 into 10 raised power
05:00
minus power minus 27. When we use, we have 1130.22 to divide 3.34 divided by 1.34 into 1130.22. So,
05:02
When we have 1130.22 to 3.314 divide, then we have answer.
05:13
Enter.
05:24
So, root mean square speed will be equal to.
05:53
Under root 338.389 into 10 raised power minus 23 plus 27.
06:16
So, it will be 4.
06:19
The under root of 338.389 is 18.395 into.
06:34
The under root of 10 raised power 4 is 10 raised power 2 meter per second.
06:41
Further, we will simplify this.
06:44
Root mean square speed will be equal to.
06:47
10 raised power 2 is also equal to 100.
06:50
So, when we multiply 100 in values, this will become 18.39.5 meter per second.
07:02
So, this is our answer by using first method.
07:09
Now, second method.
07:24
We can also calculate root mean square speed by using formula.
07:31
So, root mean square speed is equal to under root 3p upon rho means density.
07:46
The pressure is obtained is 1.01 into 10 raised power 5 Pascal divided by density.
08:01
Density is also given which is 8.99 into 10 raised power minus 2.
08:08
So, 3.03 divided by 8.99 into 10 raised power minus 2.
08:30
So, 3.03 divided by 8.99, it will be under root 0.33704.
08:47
So, root mean square speed will be equal to under root.
09:15
equal to under root 0.33704 into 5 plus 2, it will be 7. So, also we can write this as
09:35
3.3704 into 10 is power 6. So, root mean square speeds will be equal to is the under root of
09:58
3.3704 is 1.8358 into the under root of 10 raised power 6 is 10 raised power 3
10:19
meter per second. So, further हम इसको simplify करेंगे 10 cube जो है
10:30
वो 1000 के equal होता है तो 1000 जब हम इस value के साथ multiply करेंगे
10:35
तो हमें root mean square speed का जो final answer मिलेगा that will be
10:40
1835.8 meter per second. And we can also write it as
10:53
जब हम इसको round off करेंगे तो हमें final answer मिलेगा
11:03
1836 meter per second. So, this is our answer by second method and this answer by first
11:17
method. हमारे संदु टेक्स्ट के बुक में answer है वो यह है. So,
11:23
इस method को हम preference देंगे to calculate root mean square speed in this numerical.
11:30
उंगे तो में एक पुझा तो हमने का उन्साद करेंगे, तुझार,
11:37
वह है कि जब हम इस दोने है को आप में आपक तो बुका के रिए
11:38
तो हमारा सामेंस, रिए सब्बचल, भीन बपना चाहते है, जिए तो हमारा रावेशोय,
Be the first to comment
Add your comment
Recommended
3:42
|
Up next
Calculate the volume occupied by a gram mole of a gas at 0 zero degree centigrade and a pressure of 1.0 atmosphere
Physics with Akhtiar Ali Bullo
6 days ago
2:25
Photoelectric Effect -04- Threshold Frequency
Rebiaz Studio
3 weeks ago
6:57
The molar mass of nitrogen gas N2 is 28 g/mol. For 100g of nitrogen, calculate the number of moles, the volume occupied at room temperature 20 degree celsius and pressure of 1.01×10^5 pa
Physics with Akhtiar Ali Bullo
3 days ago
5:18
A sample of a gas contains 3×10^24 atoms. Calculate the volume of the gas at a temperature of 300k and a pressure of 120 kPa
Physics with Akhtiar Ali Bullo
31 minutes ago
5:15
The boiling point of liquid nitrogen is -321 degree fahrenheit. Change it into equivalent kelvin temperature.
Physics with Akhtiar Ali Bullo
1 week ago
4:42
An air storage tank whose volume is 112 litres contain 3kg of air at a pressure of 18 atmosphere. How much air would have to be forced into the tank to increase the pressure to 21 atmospheres, assuming no change in temperature
Physics with Akhtiar Ali Bullo
5 days ago
7:00
The freezing point of mercury is -39 degree centigrade. Convert it into degree fahrenheit and comfort level of temperature 20 degree Celsius into kelvin
Physics with Akhtiar Ali Bullo
1 week ago
4:55
A balloon contains 0.04 m3 of air at a pressure of 120kPa. Calculate the pressure required to reduce its volume to 0.025 m3 at constant temperature.
Physics with Akhtiar Ali Bullo
4 days ago
1:23
Ideal Gas Equation Crash Course Chemistry #12 | Derivation of pv=nrt | Boyle’s Law and Charles’s Law
Shubham Kola
3 years ago
6:31
Determine the gravitational attraction between the proton and the electron in a hydrogen atom,assuming that the electron describes a circular orbit with a radius of 0.53×10⁻¹⁰ m.mass of electron=9.1×10⁻³¹ kg,mass of proton=1.67×10⁻²⁷ kg
Physics with Akhtiar Ali Bullo
5 months ago
6:24
The mass of the planet Jupiter is 1.9×10 power 27 kg and that of the sun is 2×10 power 30 kg.If the average distance between them is 7.8×10 power 11m.Find the gravitational force of the sun on Jupiter
Physics with Akhtiar Ali Bullo
5 months ago
3:20
Prove that one kilo watt hour is equal to 3.6×10 power 6 J
Physics with Akhtiar Ali Bullo
1 year ago
3:18
Calculate The Fluid Density Using Archimedes Principle | Fluid Statics | 11th Class Physics
Rebiaz Studio
2 years ago
5:03
Compute gravitational acceleration at the surface of the planet Jupiter which has a diameter as 11 times as compared with that of the earth and a mass equal to 318 times that of earth
Physics with Akhtiar Ali Bullo
5 months ago
1:26
The battery of a pocket calculator supplies 0.35A at a potential difference of 6V. what is the power rating of the calculator ?
Physics with Akhtiar Ali Bullo
3 months ago
2:04
Prove that one joule is equal to 10 power 7 erg_-_Prove that 1 J=10 power 7 erg
Physics with Akhtiar Ali Bullo
1 year ago
4:57
The torque of force F=(2i-3j+4k)N acting at the point r=(3i+2j+3k)m about the origin be
Physics with Akhtiar Ali Bullo
3 months ago
2:07
Prove that one erg is equal to 10 power -7 joule_-_ prove that 1 erg=10-7J
Physics with Akhtiar Ali Bullo
1 year ago
5:27
How to convert value of gravitational constant G from si 6.67×10 power -11 Nm2kg-2 to cgs system
Physics with Akhtiar Ali Bullo
11 months ago
5:55
Problem No 20.1 : A hydrogen atom is in its ground state . using Bohr's theory ......
Punjab Group Of Colleges
10 years ago
1:38
Gas Laws: Boyle's Law, Charles's Law, Gay-Lussac’s Law and Avogadro's Law | 12th Chemistry Lecture
Shubham Kola
3 years ago
6:44
A 10kg mass is at a distance of 1m from a 100kg mass.Find the gravitational force of attraction when a) 10kg mass exerts force on the 100kg mass b) 100 kg mass exerts force on the 10kg mass
Physics with Akhtiar Ali Bullo
5 months ago
6:41
Real Gases- Using the Van der Waals Equation
Noise Education TV
4 years ago
6:30
The Thermochemical Equation. Using Constant Pressure Calorimeter | 11th Class Chemistry
Rebiaz Studio
2 years ago
1:17
The universal gravitational constant G is equal to_Value of capital G is
Physics with Akhtiar Ali Bullo
6 months ago
Be the first to comment