Entropy Change in Isochoric Process
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Q. The entropy change when two moles of ideal monoatomic gas is heated from 200oC to 300oC reversibly and isochorically is:
- 52R ln (573273)
- 32R ln (300200)
- 3R ln (573473)
- 32R ln (573473)
Q. Which of the following is the correct expression for ΔS in case of isochoric process?
Q. The entropy change when two moles of ideal monoatomic gas is heated from 200 to 300∘C reversibly and isochorically ?
- 32Rln[300200]
- 52Rln[573273]
- 3Rln[573473]
- 32Rln[573473]
Q. The entropy change when two moles of ideal monoatomic gas is heated from 200oC to 300oC reversibly and isochorically is:
- 32R ln (300200)
- 52R ln (573273)
- 3R ln (573473)
- 32R ln (573473)
Q. CP−CV for an ideal gas is equal to R. If true enter 1 else 0
Q. The entropy change when two moles of ideal monoatomic gas is heated from 200∘C to 300∘C reversibly and isochorically?
- 32R ln(300200)
- 52R ln(573273)
- 3 R ln(573473)
- 32R ln(573473)
Q. The entropy change when two moles of ideal monoatomic gas is heated from 200∘C to 300∘C reversibly and isochorically?
- 32R ln(300200)
- 52R ln(573273)
- 3 R ln(573473)
- 32R ln(573473)
Q. Which of the following is the correct expression for ΔS in case of isochoric process?
ΔS=nCp lnT2T1
ΔS=nCp lnT1T2
ΔS=nCv lnT2T1
ΔS=nCv lnT1T2
Q. The entropy change when two moles of ideal monoatomic gas is heated from 200oC to 300oC reversibly and isochorically is:
- 32R ln (300200)
- 52R ln (573273)
- 3R ln (573473)
- 32R ln (573473)
Q. Calculate the total entropy change for the transition at 368 K of 1 mol of sulphur from the monoclinic to the rhombic solid state, is aH=−401.7 J mol1 for the transition. Assume the surroundings to be an ice-water both at OoC.
- −1.09JK−1
- −0.5JK−1
- None of these
- −0.385JK−1
Q. The entropy change when two moles of an ideal monoatomic gas is heated from 200∘C to 300∘C reversibly and isochorically?
- 3Rln(573473)
- 32Rln(573473)
- 32Rln(300200)
- 52Rln(573273)
Q. Which of the following is the correct expression for ΔS in case of isochoric process?
ΔS=nCp lnT2T1
ΔS=nCp lnT1T2
ΔS=nCv lnT2T1
ΔS=nCv lnT1T2