$C_P-C_V=R$. This R is :

1. Change in K.E.

2. Change in rotational energy

3. work done which system can do on expanding the gas per mol per degree increase in temperature

4. All correct

Heat of neutralisation of oxalic acid is -25.4 K cal mol^-1 using strong base, NaOH. Hence enthalpy change of the process is $H_2{C}_2{O}_4\rightleftharpoons 2H^{+}+C_2{H_4}^{2-}$is-

1. 2.0 kcal

2. -11.8 kcal

3. 1.0 kcal

4. -1.0 kcal

5 moles of nitrogen gas are enclosed in an adiabatic cylindrical vessel. The piston itself is a rigid light cylindrical container containing 3 moles of Helium gas. There is a heater which gives out a power 100 cal to the nitrogen gas. A power of 30 cal is transferred to Helium through the bottom surface of the piston.

The rate of increment of temperature of the nitrogen gas assuming that the piston moves slowly :

1. 2K/sec

2. 4K/sec

3. 6K/sec

4. 8K/sec

During the isothermal mixing of ideal gases at pressure, p, the entropy change per mole for the mixing process is-R $\sum x_i \ln x_i$ where x₁, x₂,....,x_i are the mole fractions of the components, 1, 2,....,i of the mixture. Assuming ideal gas behavior, calculate $∆S$ for the mixing of 0.8 mole of N₂ and 0.2 mole of O₂.(at ${25}^{\circ }C$ and 0.9 atm) [1 eu = cal/deg]

1. 0.9943 eu

2. 0.7533 eu

3. 0.6798 eu

4. 0.7112 eu

An ideal gas can be expanded from an initial state to a certain volume through two different processes, (A) $P{V}^2=K$ (B) $P=K{V}^2$, where K is a positive constant. Then, choose the correct option from the following.

1. Final temperature in (A) will be greater than in (B)

2. Final temperature in (B) will be greater than in (A)

3. Work done by the gas in both the processes would be equal

4. Total heat given to the gas in (A) is greater than in (B)

Heat of hydrogenation of ethene is x₁ and that of benzene is x₂. Hence resonance energy is-

1. $x_1-x_2$

2. $x_1+x_2$

3. $3x_1-x_2$

4. $x_1-3x_2$

$$\begin{gathered} {\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)+3.5{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) \\ ∆{\mathrm{S}}_{\mathrm{vap}}\left({\mathrm{H}}_2\mathrm{O}, \mathrm{l}\right)={\mathrm{x}}_1 \mathrm{cal} {\mathrm{K}}^{-1}\left(\mathrm{b}.\mathrm{p}. +{\mathrm{T}}_1\right) \\ ∆{\mathrm{H}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}, \mathrm{l}\right)={\mathrm{x}}_2; ∆{\mathrm{H}}_{\mathrm{f}}\left({\mathrm{CO}}_2\right)={\mathrm{x}}_3, ∆{\mathrm{H}}_{\mathrm{f}}\left({\mathrm{C}}_2{\mathrm{H}}_6\right)={\mathrm{x}}_4 \\ \mathrm{Hence} ∆\mathrm{H} \mathrm{for} \mathrm{the} \mathrm{reaction} \mathrm{is}- \end{gathered}$$

1. $2x_3+3x_2-x_4$

2. $2x_3+3x_2-x_4+3x_1{T}_1$

3. $2x_3+3x_2-x_4-3x_1{T}_1$

4. $x_1{T}_1+x_2+x_3-x_4$

The bond energies of $\mathrm{C}\equiv \mathrm{C}$, C-H, H-H, and C=C are 198, 98, 103, and 145 kcal respectively.

The enthalpy change of the reaction $\mathrm{HC}\equiv \mathrm{CH}+{\mathrm{H}}_2\to {\mathrm{C}}_2{\mathrm{H}}_4$ would be:

1. 48 kcal

2. 96 kcal

3. -40 kcal

4. -152 kcal

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)$

B.E. (H-H) = x₁; B.E. (O=O) = x₂ B.E. (O-H) = x₃

Latent heat of vaporization of water liquid into water vapour = x₄, then $∆H_f$(heat of formation of liquid water) is-

1. $x_1+\frac{x_2}{2}-x_3+x_4$

2. $2x_3-x_1-\frac{x_2}{2}-x_4$

3. $x_1+\frac{x_2}{2}-2x_3-x_4$

4. $x_1+\frac{x_2}{2}-2x_3+x_4$

In a process the pressure of a gas is inversely proportional to the square of the volume. If temperature of the gas is increased, then work done on the gas-

1. is positive

2. is negative

3. is zero

4. maybe positive