An ideal gas goes from A to B via two processes, l and ll, as shown. If $∆{\mathrm{U}}_1$ and $∆{\mathrm{U}}_2$ are the changes in internal energies in processes I and II, respectively, then (\(P:\) pressure, \(V:\) volume)

If a gas changes volume from 2 litres to 10 litres at a constant temperature of 300K, then the change in its internal energy will be:

The incorrect relation is:
(where symbols have their usual meanings)

1. ${\mathrm{C}}_{\mathrm{P}}=\frac{\mathrm{\gamma R}}{\mathrm{\gamma }-1}$

2. ${\mathrm{C}}_{\mathrm{P}}-{\mathrm{C}}_{\mathrm{V}}=\mathrm{R}$

3. $∆\mathrm{U}=\frac{{\mathrm{P}}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{f}}-{\mathrm{P}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}}{1-\mathrm{\gamma }}$

4. ${\mathrm{C}}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{\gamma }-1}$

If 3 moles of a monoatomic gas do 150 J of work when it expands isobarically, then a change in its internal energy will be:

If n moles of an ideal gas is heated at a constant pressure from 50°C to 100°C, the increase in the internal energy of the gas will be: \(\left(\frac{C_{p}}{C_{v}} = \gamma\   and\   R = gas\   constant\right)\)

In the P-V graph shown for an ideal diatomic gas, the change in the internal energy is:
                                
1. \(\frac{3}{2}P(V_2-V_1)\)
2. \(\frac{5}{2}P(V_2-V_1)\)
3. \(\frac{3}{2}P(V_1-V_2)\)
4. \(\frac{7}{2}P(V_1-V_2)\)

If the ratio of specific heat of a gas at constant pressure to that at constant volume is $\mathrm{\gamma }$, the change in internal energy of a mass of gas, when the volume changes from V to 2V at constant pressure, P is:

1. $\mathrm{R}/\left(\mathrm{\gamma }-1\right)$

2. PV

3. $\mathrm{PV}/\left(\mathrm{\gamma }-1\right)$

4. $\mathrm{PV}\left(\mathrm{\gamma }-1\right)$

When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is?

The pressure in a monoatomic gas increases linearly from 4 atm to 8 atm when its volume increases from 0.2 m${}^3$ to 0.5 m${}^3$. The increase in internal energy will be:

If 32 gm of \(O_2\)${\mathrm{}}_{}$ at \(27^{\circ}\mathrm{C}\) is mixed with 64 gm of \(O_2\)${\mathrm{}}_{}$ at \(327^{\circ}\mathrm{C}\) in an adiabatic vessel, then the final temperature of the mixture will be:
1. \(200^{\circ}\mathrm{C}\)
2. \(227^{\circ}\mathrm{C}\)
3. \(314.5^{\circ}\mathrm{C}\)
4. \(235.5^{\circ}\mathrm{C}\)$\mathrm{}$