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Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2 at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2 in flask 1 and flask 2 are:
  • a)
    Moles in flask 1 = 0.4, Moles in flask 2 = 0.3
  • b)
    Moles in flask 1 = 0.2, Moles in flask 2 = 0.3
  • c)
    Moles in flask 1 = 0.3, Moles in flask 2 = 0.2
  • d)
    Moles in flask1= 0.4, Moles in flask3 = 0.2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Two flasks of equal volume are connected by a narrow tube (of negligib...
PV=nRT 
0.5V=0.7R300
so V=420R
In 2nd case the pressure will be same in both flasks and sum of moles of gas will be 0.7. But volume will be half 420R/2=210R
Flask 1-
PV=aRT
P210R=aR300⇒a=0.7P
Flask2-
PV=bRT 
P210R=bR400⇒b=0.525P
a+b=1.225P=0.7
P=0.571atm
a=0.7*0.571=0.399=0.4moles
b=0.525*0.571=0.299=0.3moles
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Most Upvoted Answer
Two flasks of equal volume are connected by a narrow tube (of negligib...


P,V  = const
n1T1 = n2T2
Free Test
Community Answer
Two flasks of equal volume are connected by a narrow tube (of negligib...
Given:
- Two flasks of equal volume connected by a narrow tube
- Temperature of both flasks is 27°C
- Pressure in both flasks is 0.5 atm
- Total moles of H2 in both flasks is 0.70

To find:
- Moles of H2 in each flask after one flask is immersed in a bath at 127°C

Solution:
Step 1: Calculate the initial volume of each flask using ideal gas equation:
PV = nRT

Given:
P = 0.5 atm
V = unknown (let's assume it as 'V')
n = 0.70 moles
R = 0.0821 L.atm/mol.K
T = 27°C = 27 + 273 = 300 K

0.5V = 0.70 * 0.0821 * 300

V = (0.70 * 0.0821 * 300) / 0.5

V = 34.83 L

So, the initial volume of each flask is 34.83 L.

Step 2: After one flask is immersed in a bath at 127°C, the temperature of that flask will increase to 127°C. The other flask remains at 27°C.

Step 3: Calculate the final volume of each flask using ideal gas equation:
Given:
P = 0.5 atm
V = 34.83 L
n = unknown (let's assume it as 'n')
R = 0.0821 L.atm/mol.K
T = 127°C = 127 + 273 = 400 K

0.5 * 34.83 = n * 0.0821 * 400

n = (0.5 * 34.83) / (0.0821 * 400)

n = 0.3 moles

So, the number of moles of H2 in flask 1 is 0.3 moles and in flask 2 is 0.4 moles.

Answer:
The correct answer is option 'A':
- Moles in flask 1 = 0.4
- Moles in flask 2 = 0.3
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Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer?
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Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer?.
Solutions for Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11. Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27°C and contain 0.70 moles of H2at 0.5 atm. One of the flask is then immersed into a bath kept at 127° C, while the other remains at 27°C . The number of moles of H2in flask 1 and flask 2 are:a)Moles in flask 1 = 0.4, Moles in flask 2 = 0.3b)Moles in flask 1 = 0.2, Moles in flask 2 = 0.3c)Moles in flask 1 = 0.3, Moles in flask 2 = 0.2d)Moles in flask1= 0.4, Moles in flask3 = 0.2Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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