JEE Main 2017 (Online) 9th April Morning Slot | Thermodynamics Question 140 | Chemistry

A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A :

10 J of the work will be done by the gas.

6 J of the work will be done by the gas.

10 J of the work will be done by the surrounding on gas.

6 J of the work will be done by the surrounding on gas.

JEE Main 2017 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

For a reaction, A(g) $$ \to $$ A($$\ell $$); $$\Delta $$H= $$-$$ 3RT.
The correct statement for the reaction is :

$$\Delta $$H = $$\Delta $$U $$ \ne $$ O

$$\Delta $$H = $$\Delta $$U = O

$$\left| {} \right.$$$$\Delta $$H$$\left| {} \right.$$ < $$\left| {} \right.$$$$\Delta $$U$$\left| {} \right.$$

$$\left| {} \right.$$$$\Delta $$H$$\left| {} \right.$$ > $$\left| {} \right.$$$$\Delta $$U$$\left| {} \right.$$

JEE Main 2017 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

The enthalpy change on freezing of 1 mol of water at 5^oC to ice at −5^oC is :

(Given $$\Delta $$_fusH = 6 kJ mol^$$-$$1 at 0^oC,
C_p(H₂O, $$\ell $$ = 75.3J mol^$$-$$1 K^$$-$$1)
C_p(H₂O s) =36.8 J mol^$$-$$1 K^$$-$$1)

5.44 kJ mol^$$-$$1

5.81 kJ mol^$$-$$1

6.56 kJ mol^$$-$$1

6.00 kJ mol^$$-$$1

JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

$$\Delta $$U is equal to :

Isobaric work

Adiabatic work

Isothermal work

Isochoric work