The first law of thermodynamics is based on:

The latent heat of vaporisation of water is \(2240~\text{J/gm}\). If the work done in the process of expansion of \(1~\text{g}\) is \(168~\text{J}\), then the increase in internal energy is:
1. \(2408~\text{J}\)
2. \(2240~\text{J}\)
3. \(2072~\text{J}\)
4. \(1904~\text{J}\)

An ideal gas goes from state \(A\) to state \(B\) via three different processes, as indicated in the \(P\text-V\) diagram. If \(Q_1,Q_2,Q_3\)${\mathrm{}}_{}$ indicates the heat absorbed by the gas along the three processes and \(\Delta U_1, \Delta U_2, \Delta U_3\) indicates the change in internal energy along the three processes respectively, then:

1 kg of gas does 20 kJ of work and receives 16 kJ of heat when it is expanded between two states. The second kind of expansion can be found between the same initial and final states, which requires a heat input of 9 kJ. The work done by the gas in the second expansion will be:

A sample of \(0.1\) g of water at \(100^{\circ}\mathrm{C}\)$$ and normal pressure (\(1.013 \times10^5\) N m^–2) requires \(54\) cal of heat energy to convert it into steam at \(100^{\circ}\mathrm{C}\)$$. If the volume of the steam produced is \(167.1\) cc, then the change in internal energy of the sample will be:
1. \(104.3\) J
2. \(208.7\) J
3. \(42.2\) J
4. \(84.5\) J

In an adiabatic expansion of a gas, if the initial and final temperatures are \(T_1\) and \(T_2\), respectively, then the change in internal energy of the gas is:
1. \(\frac{nR}{\gamma-1}(T_2-T_1)\)
2. \(\frac{nR}{\gamma-1}(T_1-T_2)\)
3. \(nR ~(T_1-T_2)\)
4. Zero

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. \(3P_0V_0+ 2P_0V_0\ln 2\)