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One mole of an ideal diatomic gas Cv=5cal was transformed from initial 25°C and 1L to the state when temperature is 100°C and volume 10L. The entropy change of the process can be expressed as (R=2 calories/mol/K)

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Important Questions on Thermodynamics

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For the following reaction,  CDiamond +O2(g)  CO2(g) ; ΔH=-94.0kcal
CGraphite +O2(g)CO2(g) ; ΔH=-97.6kcal
The heat réquired to change 1g of C, Cgraphite  is
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C(s)+O2(g)CO2,(g) ;    ΔH=-94.3 kcal/mol

CO(g)+12O2(g)CO2(g) ;  ΔH=-67.4 kcal/mol

O2(g)2O(g) ;  ΔH=117.4 kcal/mol

CO(g)C(g)+O(g) ; ΔH=230.6 kcal/mol
Calculate ΔH for C (s) C(g) in kcal/mol.

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During isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 273 K, the work done
is : (gas constant = 2)
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One mole of an ideal gas for which Cv=3/2R is heated reversibly at a constant pressure of 1 atm from
25°C to 100°C. The ΔH is
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The heat of reaction for

C10H8s+12O2g10CO2g+4H2O(l) at constant volume is -1228.2 kcal at 250C. The heat of reaction at constant pressure and same temperature is

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Two moles of an ideal gas are compressed isothermally 100°C and reversibly from a pressure of 10 atm to 25 atm, then the free energy change is:
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One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temprature of 27°C. The work done is 3 kJ. The final temprature of the gas is equal to [Cv=20 kJ-1] F
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Internal energy does not include: