The first law of thermodynamics is based on:
1.  the concept of temperature. 
2.  the concept of conservation of energy. 
3.  the concept of working of heat engine. 
4.  the concept of entropy. 
Two cylinders, A and B, of equal capacity are connected to each other via a stopcock. A contains gas at a standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. If the stopcock is suddenly opened, then the change in internal energy of the gas is:
1.  0

2.  5 J

3.  1 J

4.  3 J 
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. \(1904~\text{J}\)
When a system is moved from state a to state b along the path acb, it is discovered that the system absorbs 200 J of heat and performs 80 J of work. Along the path adb, heat absorbed Q = 144 J. The work done along the path adb is:
1.  6 J  2.  12 J 
3.  18 J  4.  24 J 
We consider a thermodynamic system. If ∆U represents the increase in its internal energy and W the work done by the system, which of the following statements is true?
1.  \(\Delta \mathrm{U}=\mathrm{W}\) in an isothermal process 
2.  \(\Delta \mathrm{U}=\mathrm{W}\) in an isothermal process 
3.  \(\Delta \mathrm{U}=\mathrm{W}\) in an adiabatic process 
4.  \(\Delta \mathrm{U}=\mathrm{W}\) in an adiabatic process 
A system is taken from state A to state B along two different paths, 1 and 2. If the heat absorbed and work done by the system along these two paths are ${Q}_{1},$ ${Q}_{2}$ $and$ ${W}_{1},$ ${W}_{2}$ respectively, then:
1.  \(Q_1=Q_2\) 
2.  \(W_1=W_2\) 
3.  \(Q_1W_1=Q_2W_2\) 
4.  \(Q_1+W_1=Q_2+W_2\) 
An ideal gas goes from state A to state B via three different processes, as indicated in the PV diagram. If ${\mathrm{Q}}_{1},$ ${\mathrm{Q}}_{2},$ ${\mathrm{Q}}_{3}$ indicates the heat absorbed by the gas along the three processes and $\u2206{\mathrm{U}}_{1},$ $\u2206{\mathrm{U}}_{2},$ $\u2206{\mathrm{U}}_{3}$ indicates the change in internal energy along the three processes respectively, then:
1. \(\mathrm{Q}_1>\mathrm{Q}_2>\mathrm{Q}_3 \) and \(\Delta \mathrm{U}_1=\Delta \mathrm{U}_2=\Delta \mathrm{U}_3\)
2. \(\mathrm{Q}_3>\mathrm{Q}_2>\mathrm{Q}_1\) and \(\Delta \mathrm{U}_1=\Delta \mathrm{U}_2=\Delta \mathrm{U}_3\)
3. \(\mathrm{Q}_1=\mathrm{Q}_2=\mathrm{Q}_3\) and \(\Delta \mathrm{U}_1>\Delta \mathrm{U}_2>\Delta \mathrm{U}_3\)
4. \(\mathrm{Q}_3>\mathrm{Q}_2>\mathrm{Q}_1\) and \(\Delta \mathrm{U}_1>\Delta \mathrm{U}_2>\Delta \mathrm{U}_3\)
An ideal monoatomic gas \(\left(\gamma = \frac{5}{3}\right )\) absorbs 50 cal in an isochoric process. The increase in internal energy of the gas is:
1.  20 cal  2.  Zero 
3.  50 cal  4.  30 cal 
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:
1.  32 kJ  2.  5 kJ 
3.  4 kJ  4.  13 kJ 
The work done by an ideal diatomic gas in its sudden expansion is 20 J. The change in the internal energy of the gas will be:
1. 20 J
2. 0 J
3. $20$ J
4. $15$ J