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One mole of a substance is heated from 300 K to 400 K at constant pressure. The CP of the sub stance is given by cp(J-1mol-1)= 5+0.1 T. the change in entropy, in JK-1mol-1, of the substance is ______________
Correct answer is between '11.3,11.5'. Can you explain this answer?
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One mole of a substance is heated from 300 K to 400 K at constant pres...
Calculation of Change in Entropy of a Substance
Given data:
- One mole of substance is heated from 300 K to 400 K at constant pressure.
- The CP of the substance is given by cp(J-1mol-1) = 50.1 T.
To calculate the change in entropy, we need to use the formula:
∆S = ∫(cp/T)dT
where,
∆S = change in entropy (in JK-1mol-1)
cp = specific heat capacity at constant pressure (in J-1mol-1K-1)
T = temperature (in K)
Step 1: Integrate the given CP equation
cp = 50.1 T
∫(cp/T)dT = ∫(50.1)d(lnT) = 50.1 ln(T) + C
where C is the constant of integration
Step 2: Evaluate the constant of integration
At T = 300 K, cp = 50.1 x 300 = 15030 J-1mol-1K-1
Using this value in the integrated equation,
15030 = 50.1 ln(300) + C
C = 15030 - 50.1 ln(300)
Step 3: Calculate the change in entropy
Using the integrated equation and the constant of integration, we can now calculate the change in entropy (∆S) as:
∆S = [50.1 ln(400) - 50.1 ln(300)] = 11.4 JK-1mol-1
Therefore, the change in entropy of the substance is 11.4 JK-1mol-1, which is between the given range of 11.3 and 11.5 JK-1mol-1.
Given data:
- One mole of substance is heated from 300 K to 400 K at constant pressure.
- The CP of the substance is given by cp(J-1mol-1) = 50.1 T.
To calculate the change in entropy, we need to use the formula:
∆S = ∫(cp/T)dT
where,
∆S = change in entropy (in JK-1mol-1)
cp = specific heat capacity at constant pressure (in J-1mol-1K-1)
T = temperature (in K)
Step 1: Integrate the given CP equation
cp = 50.1 T
∫(cp/T)dT = ∫(50.1)d(lnT) = 50.1 ln(T) + C
where C is the constant of integration
Step 2: Evaluate the constant of integration
At T = 300 K, cp = 50.1 x 300 = 15030 J-1mol-1K-1
Using this value in the integrated equation,
15030 = 50.1 ln(300) + C
C = 15030 - 50.1 ln(300)
Step 3: Calculate the change in entropy
Using the integrated equation and the constant of integration, we can now calculate the change in entropy (∆S) as:
∆S = [50.1 ln(400) - 50.1 ln(300)] = 11.4 JK-1mol-1
Therefore, the change in entropy of the substance is 11.4 JK-1mol-1, which is between the given range of 11.3 and 11.5 JK-1mol-1.
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One mole of a substance is heated from 300 K to 400 K at constant pres...
As all we know
H=U+PV
dH=dU+PdV+VdP (dirac(Q) =dU+PdV=TdS)
dH=TdS+VdP
constant pressure is given
above equation reduced to
CpdT=TdS
put value of Cp
[(5+0.1T) /T]*dT=dS
integrating both side
[5*ln(T) +0.1T]=[S]
putting lower upper limit of temp the answer is
11.43J/(mol*K)
H=U+PV
dH=dU+PdV+VdP (dirac(Q) =dU+PdV=TdS)
dH=TdS+VdP
constant pressure is given
above equation reduced to
CpdT=TdS
put value of Cp
[(5+0.1T) /T]*dT=dS
integrating both side
[5*ln(T) +0.1T]=[S]
putting lower upper limit of temp the answer is
11.43J/(mol*K)
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One mole of a substance is heated from 300 K to 400 K at constant pressure. The CPof the sub stance is given by cp(J-1mol-1)= 5+0.1 T. the change in entropy, in JK-1mol-1, of the substance is ______________Correct answer is between '11.3,11.5'. Can you explain this answer?
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One mole of a substance is heated from 300 K to 400 K at constant pressure. The CPof the sub stance is given by cp(J-1mol-1)= 5+0.1 T. the change in entropy, in JK-1mol-1, of the substance is ______________Correct answer is between '11.3,11.5'. 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 One mole of a substance is heated from 300 K to 400 K at constant pressure. The CPof the sub stance is given by cp(J-1mol-1)= 5+0.1 T. the change in entropy, in JK-1mol-1, of the substance is ______________Correct answer is between '11.3,11.5'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of a substance is heated from 300 K to 400 K at constant pressure. The CPof the sub stance is given by cp(J-1mol-1)= 5+0.1 T. the change in entropy, in JK-1mol-1, of the substance is ______________Correct answer is between '11.3,11.5'. Can you explain this answer?.
One mole of a substance is heated from 300 K to 400 K at constant pressure. The CPof the sub stance is given by cp(J-1mol-1)= 5+0.1 T. the change in entropy, in JK-1mol-1, of the substance is ______________Correct answer is between '11.3,11.5'. 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 One mole of a substance is heated from 300 K to 400 K at constant pressure. The CPof the sub stance is given by cp(J-1mol-1)= 5+0.1 T. the change in entropy, in JK-1mol-1, of the substance is ______________Correct answer is between '11.3,11.5'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of a substance is heated from 300 K to 400 K at constant pressure. The CPof the sub stance is given by cp(J-1mol-1)= 5+0.1 T. the change in entropy, in JK-1mol-1, of the substance is ______________Correct answer is between '11.3,11.5'. Can you explain this answer?.
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