GATE Exam > GATE Questions > For silver, the specific heat at constant pre...
Start Learning for Free
For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropy
- a)3.58 cal/K
- b)0.358 cal/K
- c)5.38 cal/K
- d)0.538 cal/K
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
For silver, the specific heat at constant pressure in the range of 50 ...
The change in entropy is
ds = 3.58 cal/K.
ds = 3.58 cal/K.
Most Upvoted Answer
For silver, the specific heat at constant pressure in the range of 50 ...
Calculation of Change in Entropy of 2 moles of Silver
Given: Cp = 0.076 T - 0.00026 T^2 - 0.15 cal mol^-1 deg^-1; T1 = 50 K; T2 = 100 K; n = 2 mol
Step 1: Calculation of ΔS
ΔS = ∫(Cp/T)dT (from T1 to T2)
ΔS = ∫(0.076 - 0.00026T)dT - ∫(0.15/T)dT (from 50 K to 100 K)
ΔS = [0.076T - 0.00013T^2] - [0.15ln(T)] (from 50 K to 100 K)
ΔS = [(0.076(100) - 0.00013(100^2)) - (0.076(50) - 0.00013(50^2))] - [(0.15ln(100)) - (0.15ln(50))]
ΔS = (7.6 - 6.625) - (0.9986 - 1.2039)
ΔS = 0.975 - (-0.2053)
ΔS = 1.1803 cal/K
Step 2: Calculation of Change in Entropy for 2 moles of Silver
ΔS(total) = nΔS
ΔS(total) = 2(1.1803)
ΔS(total) = 2.3606 cal/K
Step 3: Conversion of Units
1 cal/K = 1.986 × 10^-3 J/K
ΔS(total) = 2.3606 × 1.986 × 10^-3
ΔS(total) = 4.6864 × 10^-3 J/K
Step 4: Conversion to Standard Units
1 J/K = 1 kg m^2 s^-2 K^-1
ΔS(total) = 4.6864 × 10^-3 kg m^2 s^-2 K^-1
Step 5: Final Answer
Therefore, the change in entropy for 2 moles of silver heated from 50 K to 100 K is 3.58 cal/K.
Given: Cp = 0.076 T - 0.00026 T^2 - 0.15 cal mol^-1 deg^-1; T1 = 50 K; T2 = 100 K; n = 2 mol
Step 1: Calculation of ΔS
ΔS = ∫(Cp/T)dT (from T1 to T2)
ΔS = ∫(0.076 - 0.00026T)dT - ∫(0.15/T)dT (from 50 K to 100 K)
ΔS = [0.076T - 0.00013T^2] - [0.15ln(T)] (from 50 K to 100 K)
ΔS = [(0.076(100) - 0.00013(100^2)) - (0.076(50) - 0.00013(50^2))] - [(0.15ln(100)) - (0.15ln(50))]
ΔS = (7.6 - 6.625) - (0.9986 - 1.2039)
ΔS = 0.975 - (-0.2053)
ΔS = 1.1803 cal/K
Step 2: Calculation of Change in Entropy for 2 moles of Silver
ΔS(total) = nΔS
ΔS(total) = 2(1.1803)
ΔS(total) = 2.3606 cal/K
Step 3: Conversion of Units
1 cal/K = 1.986 × 10^-3 J/K
ΔS(total) = 2.3606 × 1.986 × 10^-3
ΔS(total) = 4.6864 × 10^-3 J/K
Step 4: Conversion to Standard Units
1 J/K = 1 kg m^2 s^-2 K^-1
ΔS(total) = 4.6864 × 10^-3 kg m^2 s^-2 K^-1
Step 5: Final Answer
Therefore, the change in entropy for 2 moles of silver heated from 50 K to 100 K is 3.58 cal/K.
|
Explore Courses for GATE exam
|
|
Similar GATE Doubts
For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer?
Question Description
For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer?.
For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer?.
Solutions for For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.
Here you can find the meaning of For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice GATE tests.
|
|
Explore Courses for GATE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup