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One mole of an ideal gas is sub jected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K . the change in Gibbs free energy of the system is(.???)kJ/mol .( Round off to one decimal place) . (R) gas constant =8.3j/K.mol?
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One mole of an ideal gas is sub jected to an isothermal increase in pr...
Calculation of Change in Gibbs Free Energy

To determine the change in Gibbs free energy (ΔG) of the system, we can use the equation:

ΔG = ΔH - TΔS

Where:
ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
T is the temperature in Kelvin
ΔS is the change in entropy

Step 1: Calculation of ΔH

Since the gas is ideal, we can assume that it follows the ideal gas law:

PV = nRT

Where:
P is the pressure
V is the volume
n is the number of moles
R is the gas constant
T is the temperature

Given that the pressure increases from 100 kPa to 1000 kPa, and the number of moles is constant (1 mole), we can rearrange the equation to solve for the change in volume (ΔV):

ΔV = nRTΔP

Where:
ΔV is the change in volume
ΔP is the change in pressure

Substituting the values into the equation:

ΔV = (1 mol)(8.3 J/K.mol)(300 K)(1000 kPa - 100 kPa)

Simplifying the equation:

ΔV = 1 mol(8.3 J/K.mol)(300 K)(900 kPa)

Converting kilopascals (kPa) to pascals (Pa):

ΔV = 1 mol(8.3 J/K.mol)(300 K)(900,000 Pa)

Calculating the change in enthalpy (ΔH) using the equation:

ΔH = ΔU + PΔV

Where:
ΔU is the change in internal energy
P is the constant pressure

Assuming the process is isothermal, the change in internal energy (ΔU) is zero. Thus:

ΔH = PΔV

ΔH = (1000 kPa)(1 mol)(8.3 J/K.mol)(300 K)(900,000 Pa)

Step 2: Calculation of ΔS

Since the process is isothermal, the change in entropy (ΔS) can be calculated using the equation:

ΔS = nRln(P2/P1)

Where:
P2 is the final pressure
P1 is the initial pressure

Substituting the values into the equation:

ΔS = (1 mol)(8.3 J/K.mol)ln(1000 kPa/100 kPa)

Simplifying the equation:

ΔS = (1 mol)(8.3 J/K.mol)ln(10)

ΔS = (1 mol)(8.3 J/K.mol)(2.3026)

Step 3: Calculation of ΔG

Now that we have calculated ΔH and ΔS, we can substitute these values into the equation to find the change in Gibbs free energy:

ΔG = ΔH - TΔS

ΔG = (1000 kPa)(1 mol)(8.3 J/K.mol)(300 K)(900,000 Pa) - (300 K)(1 mol)(8.3 J/K.mol)(2.3026)

Calculating the
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One mole of an ideal gas is sub jected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K . the change in Gibbs free energy of the system is(.???)kJ/mol .( Round off to one decimal place) . (R) gas constant =8.3j/K.mol?
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One mole of an ideal gas is sub jected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K . the change in Gibbs free energy of the system is(.???)kJ/mol .( Round off to one decimal place) . (R) gas constant =8.3j/K.mol? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about One mole of an ideal gas is sub jected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K . the change in Gibbs free energy of the system is(.???)kJ/mol .( Round off to one decimal place) . (R) gas constant =8.3j/K.mol? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal gas is sub jected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K . the change in Gibbs free energy of the system is(.???)kJ/mol .( Round off to one decimal place) . (R) gas constant =8.3j/K.mol?.
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