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Mass of gas is 300 gm and its specific heat at constant volume is 750J/kg K. if gas is heated through 75°C at constant pressure of 105 N/m2, it expands by volume 0.08 × 106 cm3. find CP/CV.
- a)1.4
- b)1.374
- c)1.474
- d)1.5
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Mass of gas is300 gm and its specific heat at constant volume is 750J/...
Data given;
ΔT = 75 oC = 75 K
ΔV = 0.08 * 106 cm3 = 0.08 m3
Cv = 750 J kg-1 K-1
m = 300 g = 0.300 kg
p = 105 N m-2 = 100000 N m-2
The first law of thermodynamics;
ΔU = Q – W
Q = ΔU + W ---- (1)
mCpΔT = mCvΔT + pΔV ----- (2)
Dividing the above expression (2) with Cv , we get;
Cp / Cv = Cv / Cv + pΔV / mΔTCv
Cp / Cv = 1 + pΔV / mΔTCv
Cp / Cv = 1 + ([100000 N m-2 . 0.08 m3] / [0.300 kg . 75 K .750 J kg-1K-1])
Cp / Cv = 1 + 0.474
Cp / Cv = 1.474 ; After rounding it off we get,
= 1.5 (appx)
ΔT = 75 oC = 75 K
ΔV = 0.08 * 106 cm3 = 0.08 m3
Cv = 750 J kg-1 K-1
m = 300 g = 0.300 kg
p = 105 N m-2 = 100000 N m-2
The first law of thermodynamics;
ΔU = Q – W
Q = ΔU + W ---- (1)
mCpΔT = mCvΔT + pΔV ----- (2)
Dividing the above expression (2) with Cv , we get;
Cp / Cv = Cv / Cv + pΔV / mΔTCv
Cp / Cv = 1 + pΔV / mΔTCv
Cp / Cv = 1 + ([100000 N m-2 . 0.08 m3] / [0.300 kg . 75 K .750 J kg-1K-1])
Cp / Cv = 1 + 0.474
Cp / Cv = 1.474 ; After rounding it off we get,
= 1.5 (appx)
Most Upvoted Answer
Mass of gas is300 gm and its specific heat at constant volume is 750J/...
Cp= cv + R;;
MCp ∆ t = MCv ∆t + MR∆t
M is mass
P∆v = MR ∆t
so cp/cv = 1+ p ∆ v/ MCv∆t
Take mass in kg M ,∆ t = 75 only not in Kelvin coz it is ∆ t ,
p in Pascal ,nd so v in m^3
MCp ∆ t = MCv ∆t + MR∆t
M is mass
P∆v = MR ∆t
so cp/cv = 1+ p ∆ v/ MCv∆t
Take mass in kg M ,∆ t = 75 only not in Kelvin coz it is ∆ t ,
p in Pascal ,nd so v in m^3
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Community Answer
Mass of gas is300 gm and its specific heat at constant volume is 750J/...
Data given;
ΔT = 75 oC = 75 K
ΔV = 0.08 * 106 cm3 = 0.08 m3
Cv = 750 J kg-1 K-1
m = 300 g = 0.300 kg
p = 105 N m-2 = 100000 N m-2
The first law of thermodynamics;
ΔU = Q – W
Q = ΔU + W ---- (1)
mCpΔT = mCvΔT + pΔV ----- (2)
Dividing the above expression (2) with Cv , we get;
Cp / Cv = Cv / Cv + pΔV / mΔTCv
Cp / Cv = 1 + pΔV / mΔTCv
Cp / Cv = 1 + ([100000 N m-2 . 0.08 m3] / [0.300 kg . 75 K .750 J kg-1K-1])
Cp / Cv = 1 + 0.474
Cp / Cv = 1.474 ; After rounding it off we get,
= 1.5 (appx)
ΔT = 75 oC = 75 K
ΔV = 0.08 * 106 cm3 = 0.08 m3
Cv = 750 J kg-1 K-1
m = 300 g = 0.300 kg
p = 105 N m-2 = 100000 N m-2
The first law of thermodynamics;
ΔU = Q – W
Q = ΔU + W ---- (1)
mCpΔT = mCvΔT + pΔV ----- (2)
Dividing the above expression (2) with Cv , we get;
Cp / Cv = Cv / Cv + pΔV / mΔTCv
Cp / Cv = 1 + pΔV / mΔTCv
Cp / Cv = 1 + ([100000 N m-2 . 0.08 m3] / [0.300 kg . 75 K .750 J kg-1K-1])
Cp / Cv = 1 + 0.474
Cp / Cv = 1.474 ; After rounding it off we get,
= 1.5 (appx)
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Mass of gas is300 gm and its specific heat at constant volume is 750J/kg K. if gas isheated through 75°C at constantpressure of 105N/m2, it expands by volume 0.08 ×106cm3. find CP/CV.a)1.4b)1.374c)1.474d)1.5Correct answer is option 'D'. 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 Mass of gas is300 gm and its specific heat at constant volume is 750J/kg K. if gas isheated through 75°C at constantpressure of 105N/m2, it expands by volume 0.08 ×106cm3. find CP/CV.a)1.4b)1.374c)1.474d)1.5Correct answer is option 'D'. 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 Mass of gas is300 gm and its specific heat at constant volume is 750J/kg K. if gas isheated through 75°C at constantpressure of 105N/m2, it expands by volume 0.08 ×106cm3. find CP/CV.a)1.4b)1.374c)1.474d)1.5Correct answer is option 'D'. Can you explain this answer?.
Mass of gas is300 gm and its specific heat at constant volume is 750J/kg K. if gas isheated through 75°C at constantpressure of 105N/m2, it expands by volume 0.08 ×106cm3. find CP/CV.a)1.4b)1.374c)1.474d)1.5Correct answer is option 'D'. 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 Mass of gas is300 gm and its specific heat at constant volume is 750J/kg K. if gas isheated through 75°C at constantpressure of 105N/m2, it expands by volume 0.08 ×106cm3. find CP/CV.a)1.4b)1.374c)1.474d)1.5Correct answer is option 'D'. 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 Mass of gas is300 gm and its specific heat at constant volume is 750J/kg K. if gas isheated through 75°C at constantpressure of 105N/m2, it expands by volume 0.08 ×106cm3. find CP/CV.a)1.4b)1.374c)1.474d)1.5Correct answer is option 'D'. Can you explain this answer?.
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