Thermodynamics ...

Heat of $$30 kcal$$ is supplied to a system and $$4200 J$$ of external work is done on the system so that  its volume decreases at constant pressure. What is the change in its internal energy. $$=(J=4200J/kcal)$$

A diatomic gas is heated at constant at constant pressure. What fraction of the heat energy is used to increase the internal energy ?

The $$\Delta H$$ and $$\Delta S$$ for a reaction at one atmospheric pressure are $$+30.558\ KJ$$ and $$+0.066\ KJ$$ respectively. The temperature at which the free energy changes will be zero and below this temperature, the nature of the reaction would be

Given that:
(i) $$C$$ (graphite)$$+O_2(g)\rightarrow CO_2(g)$$; $$\quad \Delta _rH^o=x\ kJ mol^{-1}$$.
(ii) $$C$$ (graphite)$$+\dfrac{1}{2}O_2(g)\rightarrow CO(g)$$; $$\quad \Delta _rH^o=y\ kJ mol^{-1}$$
(iii) $$CO(g)+\dfrac{1}{2}O_2(g)\rightarrow CO_2(g)$$; $$\quad \Delta _rH^o=z\ kJ mol^{-1}$$

For $$1^{st}$$ law of thermodynamics, select the correct option. 

The internal energy of gases $$He, O_{2}$$ and $$NH_{3}$$ are plotted against the absolute temperature. The respective graphs $$1, 2$$ and $$3$$ are of?

The sole criterion for the spontaneity of a process is 

Consider the following reaction
$${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\left( \ell  \right){\text{ + }}\frac{{{\text{15}}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{6C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right){\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{O}}(g)$$
Signs of $$\Delta {\text{H,}}\,\,\Delta {\text{S}}\,{\text{and}}\,\Delta {\text{G}}$$ for the above reaction will be:

At $$ 27^oC $$ the reaction,
$$ C_6H_{6(1)} + \dfrac{15}{2} O_{2(g)} \rightarrow 6CO_{2(g)} +3H_2O_{(I)} $$
proceeds spontaneously because the magnitude of 

A process has $$\Delta H = 200 \,J \,mol^{-1}$$ and $$\Delta S = 40 \,JK \,mol^{-1}$$. Out of the values given above which the process will be sponteneous: