JEE Main 2013 (Offline) | Thermodynamics Question 205 | Chemistry

A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0^oC. As it does so, it absorbs 208J of heat. The values of q and w for the process will be :
(R = 8.314 J/mol K) ( l n 7.5 = 2.01)

q = – 208 J, w = – 208 J

q = – 208 J, w = + 208 J

q = + 208 J, w = + 208 J

q = + 208 J, w = – 208 J

AIEEE 2012

MCQ (Single Correct Answer)

-1

The incorrect expression among the following is :

$${{\Delta {G_{system}}} \over {\Delta {S_{total}}}} = - T$$

In isothermal process $${w_{reversible}}$$ = $$ - nRT\,\ln \,{{{V_f}} \over {{V_i}}}$$

In $$K\, = {{\Delta {H^o} - T\Delta {S^o}} \over {RT}}$$

$$K\, = \,{e^{ - \Delta {G^o}/RT}}$$

AIEEE 2011

MCQ (Single Correct Answer)

-1

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm³ to a volume of 100 dm³ at 27^oC is :

35.8 J mol^-1 K⁻¹

32.3 J mol^-1 K⁻¹

42.3 J mol^-1 K⁻¹

38.3 J mol^-1 K⁻¹

AIEEE 2010

MCQ (Single Correct Answer)

-1

For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If T_e is the temperature at equilibrium, the reaction would be spontaneous when :

T_e > T

T > T_e

T_e is 5 times T

T = T_e

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