Work Done in Isothermal Reversible Process
Trending Questions
Match gases under specified conditions listed in Column 1 with their properties/laws in Column II.
Column IColumn IIAHydrogen gas ⟮p = 200 atm, T = 273 K⟯pCompression factor = 1BHydrogen gas ⟮p = 0 atm, T = 273 K⟯qAttractive forces are dominantCCO2⟮p = 1 atm, T = 273 K⟯rpV=nRTDReal gas with very large molar volumesp⟮V−nb⟯=nRT
A →(r), B →(p), C →(s), D →(q)
A →(p, s), B →(r), C →(p, q), D →(p, s)
A →(q), B →(p), C →(s), D →(r)
A →(p), B →(q), C →(r), D →(s)
In a process a system does 140 J of work on the surrondings and only 40 J of heat is added to the system, hence change in internal energy is
180 J
-180 J
-23.92 Cal
-180 J
- Always 100 calories per degree
- Always negative
- \N
- Always positive
- B and D
- A and D
- B and C
- A and C
- 8192.6 cal
- 9258.3 cal
- 5924.2 cal
- 2896.4 cal
a. the magnitude of work involved in an intermediate irreversible expansion is less than involved in reversible expansion.
b. the magnitude of work involved in an intermediate reversible compression is more than that involved in intermediate irreversible compression.
- Statement b
- Statement a
- neither a nor b
- A is correct while b is incorrent
- Both the statements are correct
- U and H increases
- U increases but H decreases
- U and H are unchanged
- H increases but U decreases
- Cp>>Cv
- Cp≥Cv
- Cv≥Cp
- Cv>>Cp
- △H>0 and △U=0
- △H>0 and △U<0
- △H=0 and △U=0
- △H=0 and △U>0
Calculate the work done for a reversible isothermal expansion of one mole of an ideal gas at 127∘C from a volume of 10 dm3 to 20 dm3
- -2305.3 J
- 2305. 3 J
- -2306.3 J
- 2306.3 J
- 820.8 cal
- −848.2 cal
- 84.7 cal
- −848.2 kcal
- −312 J
- +123 J
- −213 J
- +231 J
- W1>W2>W3>W4
- W3>W2>W1>W4
- W3>W2>W4>W1
- W3>W1>W2>W4
Choose the correct option from the following:
- Heat is negative
- Work done is - 965.84 cal
- All are incorrect
- Change in internal energy is positive
Isothermally at 27∘C, one mole of a vanderwaal's gas expands reversibly from 2 liters to 20 liters. Calculate the work done if a = 1.42 × 1012 dynes cm4 per mole and b = 30 c.c
- - 58.12 k J
-861.2 k J
-691.2 k J
-761.2 k J
ΔfG∘[C(graphite)]=0 kJmol−1
ΔfG∘[C(diamond)]=2.9 kJmol−1
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2×10−6m3mol−1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond) is:
[Useful information: 1 J=1 kg m2s−2; 1 Pa=1 kg m−1s−2; 1 bar=105 Pa]
- 14501 bar
- 58001 bar
- 1450 bar
- 29001 bar
- For an isothermal process, work done in reversible expansion is more than work done in irreversible expansion
- For an isothermal irreversible expansion, greater the difference in initial and final pressure, lesser is the work done
- For an isothermal irreversible compression, greater the difference in initial and final pressure, greater is the work done
- None of the above
- −1257.43 cal
- −895.8 cal
- −1499.6 cal
- −1172.6 cal
- 8.78kJ
- −1.73kJ
- 10.73kJ
- −9.78kJ
- nCvΔT
- nRγ−1(T2−T1)
- −nRPext[T2P1−T1P2P1P2]
- −2.303RTlogV2V1
- 820.8 cal
- −848.2 cal
- 84.7 cal
- −848.2 kcal
- +3.45kJ
- −8.02kJ
- +18.02kJ
- −14.01kJ
- 4.191
- 6.892
- −6.892
- −4.191
- 0, −965.84 cal
- +965.58 cal, −865.58 cal
- −965.84 cal, −865.58 cal
- +965.84 cal, +865.58 cal