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Density of an ideal gas at 298 K and 1.0 atm is found to be 1.25 kg m-3. Density of the gas at 1.5 atm and at 298 K is
- a)1.25 kg m-3
- b)1.875 kg m-3
- c)1.00 kg m-3
- d)1.20 kg m-3
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Density of an ideal gas at 298 K and1.0 atm is found to be 1.25 kg m-3...
We know ideal gas equation,
PV=nRT.
n= m(mass)/M(molecular weight).
PV=m/M RT.
PM=m/v RT.
density (d)=m(mass)/v (volume).
PM=dRT.
d=PM/RT.
here, density is directly proportional to pressure and inversely proportional to temperature...
d1/d2=P1 T2 / P2 T1.
1.25/d2= 1×298/1.5×298.
1.25/d2=1/1.5.
d2= 1.25×1.5.
d2=1.875
PV=nRT.
n= m(mass)/M(molecular weight).
PV=m/M RT.
PM=m/v RT.
density (d)=m(mass)/v (volume).
PM=dRT.
d=PM/RT.
here, density is directly proportional to pressure and inversely proportional to temperature...
d1/d2=P1 T2 / P2 T1.
1.25/d2= 1×298/1.5×298.
1.25/d2=1/1.5.
d2= 1.25×1.5.
d2=1.875
Most Upvoted Answer
Density of an ideal gas at 298 K and1.0 atm is found to be 1.25 kg m-3...
We know ideal gas equation,PV=nRT.n= m(mass)/M(molecular weight).PV=m/M RT.PM=m/v RT.density (d)=m(mass)/v (volume).PM=dRT.d=PM/RT.here, density is directly proportional to pressure and inversely proportional to temperature...d1/d2=P1 T2 / P2 T1.1.25/d2= 1×298/1.5×298.1.25/d2=1/1.5.d2= 1.25×1.5.d2=1.875....
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Community Answer
Density of an ideal gas at 298 K and1.0 atm is found to be 1.25 kg m-3...
We know ideal gas equation,
PV=nRT.
n= m(mass)/M(molecular weight).
PV=m/M RT.
PM=m/v RT.
density (d)=m(mass)/v (volume).
PM=dRT.
d=PM/RT.
here, density is directly proportional to pressure and inversely proportional to temperature...
d1/d2=P1 T2 / P2 T1.
1.25/d2= 1×298/1.5×298.
1.25/d2=1/1.5.
d2= 1.25×1.5.
d2=1.875
PV=nRT.
n= m(mass)/M(molecular weight).
PV=m/M RT.
PM=m/v RT.
density (d)=m(mass)/v (volume).
PM=dRT.
d=PM/RT.
here, density is directly proportional to pressure and inversely proportional to temperature...
d1/d2=P1 T2 / P2 T1.
1.25/d2= 1×298/1.5×298.
1.25/d2=1/1.5.
d2= 1.25×1.5.
d2=1.875
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Density of an ideal gas at 298 K and1.0 atm is found to be 1.25 kg m-3. Density ofthe gas at 1.5 atm and at 298 K isa)1.25 kg m-3b)1.875kg m-3c)1.00kg m-3d)1.20kg m-3Correct answer is option 'B'. Can you explain this answer? for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Density of an ideal gas at 298 K and1.0 atm is found to be 1.25 kg m-3. Density ofthe gas at 1.5 atm and at 298 K isa)1.25 kg m-3b)1.875kg m-3c)1.00kg m-3d)1.20kg m-3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Density of an ideal gas at 298 K and1.0 atm is found to be 1.25 kg m-3. Density ofthe gas at 1.5 atm and at 298 K isa)1.25 kg m-3b)1.875kg m-3c)1.00kg m-3d)1.20kg m-3Correct answer is option 'B'. Can you explain this answer?.
Density of an ideal gas at 298 K and1.0 atm is found to be 1.25 kg m-3. Density ofthe gas at 1.5 atm and at 298 K isa)1.25 kg m-3b)1.875kg m-3c)1.00kg m-3d)1.20kg m-3Correct answer is option 'B'. Can you explain this answer? for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Density of an ideal gas at 298 K and1.0 atm is found to be 1.25 kg m-3. Density ofthe gas at 1.5 atm and at 298 K isa)1.25 kg m-3b)1.875kg m-3c)1.00kg m-3d)1.20kg m-3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Density of an ideal gas at 298 K and1.0 atm is found to be 1.25 kg m-3. Density ofthe gas at 1.5 atm and at 298 K isa)1.25 kg m-3b)1.875kg m-3c)1.00kg m-3d)1.20kg m-3Correct answer is option 'B'. Can you explain this answer?.
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