The efficiency of an ideal heat engine is less than 100% because of:

1.	the presence of friction.
2.	the leakage of heat energy.
3.	unavailability of the sink at zero kelvin.
4.	All of these

Work done during the given cycle is:
                               
1. 4$P_0{V}_0$

2. 2$P_0{V}_0$

3. $\frac{1}{2}$$P_0{V}_0$

4. $P_0{V}_0$

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)

The pressure-temperature (P-T) graph for two processes, A and B, in a system is shown in the figure. If ${\mathrm{W}}_1$ and ${\mathrm{W}}_2$ are work done by the gas in process A and B respectively, then:

In the cyclic process shown in the pressure-volume \((P-V)\) diagram, the change in internal energy is equal to:

1. $\pi {\left(\frac{{\mathrm{P}}_2-{\mathrm{P}}_1}{2}\right)}^2$

2. $\mathrm{\pi }{\left(\frac{{\mathrm{V}}_2-{\mathrm{V}}_1}{2}\right)}^2$

3. $\frac{\mathrm{\pi }}{4}({\mathrm{P}}_2-{\mathrm{P}}_1)({\mathrm{V}}_2-{\mathrm{V}}_1)$

4. zero

             
1. \(V_1= V_2\)
2. \(V_1> V_2\)
3. \(V_1< V_2\)
4. \(V_1\ge V_2\)

\(0.04\) mole of an ideal monatomic gas is allowed to expand adiabatically so that its temperature changes from \(800~\text{K}\) to \(500~\text{K}\). The work done during expansion is nearly equal to:

The first law of thermodynamics is based on:

Find out the total heat given to diatomic gas in the process \(A\rightarrow B \rightarrow C\): \(( B\rightarrow C\) is isothermal)
                 
1. \(P_0V_0+ 2P_0V_0\ln 2\)
2. \(\frac{1}{2}P_0V_0+ 2P_0V_0\ln 2\)
3. \(\frac{5}{2}P_0V_0+ 2P_0V_0\ln 2\)
4. \(3P_0V_0+ 2P_0V_0\ln 2\)