1. A body of weight 100N falls freely a vertical distance of 50m. The atmospheric drag force is 0.5N. For the body, the work interaction is
(a) +5000 J
(b) 5000 J
(c) 25 J
(d) +25 J
(1 Mark, 1993)

Ans: d
Explanation:
A body of weight 100 N falls 50 m. It decreases the internal energy.
But the work interaction is because of the drag force.
The work by the body = Dragforce\times distancetravel=0.5\times 50 = 25 J
2. An insulated rigid vessel contains a mixture of fuel and air. The mixture is ignited by a minute spark. The contents of the vessel experience
(a) Increase in temperature, Pressure and Energy
(b) Decrease in temperature, Pressure and Energy
(c) Increase in temperature and Pressure but no change in energy
(d) Increase in temperature and Pressure but decrease in energy
(1 Mark, 1993)

Ans: c
Explanation:
After sparking the temperature will increase definitely.
If the temperature increases, then the momentum of molecules increases, the collision between the molecules and the rigid vessel increases. So, the pressure increases.
As the vessel is insulated, so no energy transfer takes place.
3. The definition of 1 K as per the internationally accepted temperature scale is
(a) \frac { 1 }{ 100 } th the difference between normal boiling point and normal freezing point of water.
(b) \frac { 1 }{ 273.15} th the normal freezing point of water
(c) 100 times the difference between the triple point of water and the normal freezing point of water.
(d) \frac { 1 }{273.16}th of the triple point of water.
(1 Mark, 1994)

Ans: d
4. The specific heats of an ideal gas depend on its
(a) Temperature
(b) Pressure
(c) Volume
(d) Molecular weight and structure
(1 Mark, 1996)

Ans: d
5. An isolated thermodynamic system executes a process. Choose the correct statement (s) from the following:
(a) No heat is transferred
(b) No work is done
(c) No mass flows across the boundary of the system
(d) No chemical reaction takes place within the system
(2 Mark, 1999)

Ans: a, b, c
6. Nitrogen at an initial state of 10 bar, 1 m^{3} and 300 K is expanded isothermally to a final volume of 2m^{3}. The pvT relation is \left( p+\frac { a }{ { v }^{ 2 } } \right) v = RT, where a > 0. the final pressure
(a) will be slightly less than 5 bar
(b) will be slightly more than 5 bar
(c) will be exactly 5 bar
(d) cannot be ascertained in the absence of the value of a
(2 Mark, 2005)

Ans: b
Explanation:
The process is isothermal.
So, { p }_{ 1 }{ v }_{ 1 }+\frac { a }{ { v }_{ 1 } } ={ p }_{ 2 }{ v }_{ 2 }+\frac { a }{ { v }_{ 2 } }
=> { 10\times 1+\frac { a }{ 1 } = }{ p }_{ 2 }\times 2+\frac { a }{ 2 }
=> { p }_{ 2 }=5+\frac { a }{ 4 }
So, the final pressure will be slightly more than 5 bar.
7. Match items from groups I, II, III, IV and V.
(a) F – G – J – K – M; E – G – I – K – N
(b) E – G – I – K – M; F – H – I – K – N
(c) F – H – J – L – N; E – H – I – L – M
(d) E – G – J – K – N; F – H – J – K – M
(2 Mark, 2006)

Ans: d
8. A gas expands in a frictionless piston cylinder arrangement. The expansion process is very slow, and is resisted by an ambient pressure of 100 kPa. During the expansion process, the pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.01 m^{3}. The maximum amount of work that could be utilized from the above process is
(a) 0 kJ
(b) 1 kJ
(c) 2 kJ
(d) 3 kJ
(2 Mark, 2008)

Ans: c
Explanation:
The work output = (p_{cyl} – p_{atm})dV = (300 – 100)0.01 = 2 kJ
8. A frictionless pistoncylinder device contains a gas initially at 0.8MPa and 0.015 m^{3}. It expands quasistatically at constant temperature to a final volume of 0.030 m^{3}. The work output (in kJ) during this process will be
(a) 8.32
(b) 12.00
(c) 554.67
(d) 8320.00
(2 Mark, 2009)

Ans: a
Explanation:
Work output for isothermal process:
W={ p }_{ 1 }V_{ 1 }ln\frac { { V }_{ 2 } }{ { V }_{ 1 } } = 800\times 0.015\times ln\frac { 0.030 }{ 0.015 } =8.32kJ
9. A monoatomic ideal gas (γ = 1.67, molecular weight = 40) is compressed adiabatically from 0.1MPa, 300K to 0.2MPa . The universal gas constant is 8.314 kJ/kgmolK. The work of compression of the gas (in kJ kg^{1}) is
(a) 29.7
(b) 19.9
(c) 13.3
(d) 0
(2 Mark, 2010)

Ans: a
Explanation:
For adiabatic process,
\frac { { T }_{ 2 } }{ { T }_{ 1 } } ={ \left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right) }^{ \frac { \gamma 1 }{ \gamma } }=> \frac { { T }_{ 2 } }{ 300 } ={ \left( \frac { 0.2 }{ 0.1 } \right) }^{ \frac { 1.671 }{ 1.67 } }
=> { T }_{ 2 }=396K
Character gas constant (R) = \frac { Universal\quad gas\quad constant }{ Molecular\quad weight } =\frac { 8.314 }{ 40 } =0.207kJ/kgK
Work done in adiabatic process is given by
W=\frac { R\left( { T }_{ 1 }{ T }_{ 2 } \right) }{ \gamma 1 } =\frac { 0.2708(300396) }{ 1.671 } =29.7kJ/kgHere, Negative sign shows the compression work.
10. A cylinder contains 5 m^{3} of an ideal gas at a pressure of 1 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 5 bar. The work in kJ required for this process is
(a) 804.7
(b) 953.2
(c) 981.7
(d) 1012.2
(1 Mark, 2013)

Ans: a
Explanation:
For Reversible isothermal process:
Work done (W)= { p }_{ 1 }{ v }_{ 1 }ln\frac { { p }_{ 1 } }{ { p }_{ 2 } }
=> W={ 10 }^{ 5 }\times 5\times ln\frac { 1 }{ 5 } =807.7kJ
11. A certain amount of an ideal gas is initially at a pressure p_{1} and temperature T_{1}. First, it undergoes a constant pressure process 12 such that T_{2} = 3T_{1}/4. Then, it undergoes a constant volume process 23 such that T_{3} = T_{1}/2. The ratio of the final volume to the initial volume of the ideal gas is
(a) 0.25
(b) 0.75
(c) 1.0
(d) 1.5
(2 Mark, 2014[3])

Ans: b
Explanation:
Process 1 – 2: Constant pressure process such that T_{2} = 3T_{1}/4
=> \frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { 3 }{ 4 }
In process 1 – 2: \frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { { v }_{ 2 } }{ { v }_{ 1 } } =\frac { 3 }{ 4 }
Process 2 – 3: Constant volume process i.e. { v }_{ 3 }={ v }_{ 2 }
Now, Ratio of final to initial volume is \frac { { v }_{ 3 } }{ { v }_{ 1 } } =\frac { { v }_{ 2 } }{ { v }_{ 1 } } =\frac { 3 }{ 4 } = 0.75
12. Which of the following statements are TRUE with respect to heat and work?
(i) They are boundary phenomena
(ii) They are exact differentials
(iii) They are path functions
(a) both (i) and (ii)
(b) both (i) and (iii)
(c) both (ii) and (iii)
(d) only (iii)
(1 Mark, 2016[1])

Ans: b
13. An ideal gas undergoes a reversible process in which the pressure varies linearly with volume. The conditions at the start (subscript 1) and at the end (subscript 2) of the process with usual notation are: p_{1}= 100 kPa, V_{1}= 0.2 m^{3} and p_{2 }= 200 kPa, V_{2}= 0.1 m^{3} and the gas constant, R = 0.275 kJ/kgK. The magnitude of the work required for the process (in kJ) is ____.
(2 Mark, 2016[1])
14. The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87 m^{3} and 300 K, respectively. The gauge pressure indicated by manometer fitted to the wall of the vessel is 0.5 bar. If the gas constant of air is R = 287 J/kgK and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel is
(a) 1.67
(b) 3.33
(c) 5.00
(d) 6.66
(2 Mark, 2017[2])

Ans: c
Explanation:
P_{abs} = P_{atm} + P_{g} = 1 + 0.5 = 1.5 bar
Using Ideal gas equation: PV = mRT
=> 150\times { 10 }^{ 3 }\times 2.87=m\times R\times 300
=> m = 5 kg
15. A mass m of a perfect gas at pressure p_{1} and volume V_{1} undergoes an isothermal process. The final pressure is p_{2} and volume is V_{2}. The work done on the system is considered positive. If R is the gas constant and T is the temperature, then the work done in the process is
(a) p_{1}V_{1}ln \frac { { v }_{ 2 } }{ v_{ 1 } }
(b) p_{1}V_{1}ln \frac { { p }_{ 1 } }{ p_{ 2 } }
(c) RTln\frac { { v }_{ 2 } }{ v_{ 1 } }
(d) mRTln \frac { { p }_{ 2 } }{ p_{ 1 } }
(1 Mark, 2017[2])

Ans: b
16. An engine operates on the reversible cycle as shown in the figure. The work output from the engine (in kJ/cycle) is ______ (correct to two decimal places).
(1 Mark, 2018[2])

Ans: 62.5
Explanation:
Work output from engine is given by the area under the curve of PV diagram.
W = \frac { 1 }{ 2 } \times (2.52)\times (650400) = 62.5 kJ/cycle
17. Air is held inside a noninsulated cylinder using a piston (mass M=25 kg and area A=100 cm^{2}) and stoppers (of negligible area), as shown in the figure. The initial pressure P_{i} and temperature T_{i} of air inside the cylinder are 200 kPa and 400^{0}C, respectively. The ambient pressure and temperature are 100 kPa and 27^{0}C, respectively. The temperature of the air inside the cylinder (^{0}C) at which the piston will begin to move is _____ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 147.625
Explanation:
The piston will just start to move when pressure on both sides is equal.
P_{atm }= 100 kPa
P_{w }= \frac { Mg }{ A } =\frac { 25\times 10 }{ { 10 }^{ 2 } } \times { 10 }^{ 3 } = 25 kPa
Pressure on outside of piston = P_{atm} + P_{w }= 100 + 25 = 125 kPa
Putting Ideal gas equation inside the cylinder:
\frac { { P }_{ i } }{ { P }_{ f } } =\frac { { T }_{ i } }{ { T }_{ f } }=> \frac { 200 }{ 125 } =\frac { 400+273 }{ { T }_{ f }}
=> { T }_{ f }=420.625K={ 147.625 }^{ 0 }C
18. If one mole of H_{2 }gas occupies a rigid container with a capacity of 1000 liters and the temperature is raised from 27°C to 37°C, the change in pressure of the contained gas (round off to two decimal places), assuming ideal gas behavior, is _____ Pa. (R = 8.314 J/mol.K).
(2 Mark, 2019[1])

Ans: 83.14 Pa
Explanation:
We know, PV = nRT
As the volume is constant: V = 1000 lit = 1 m^{3}
P_{1}V = nRT_{1}
P_{2}V = nRT_{2}
Now, (P_{2} – P_{1})V = nR(T_{2} – T_{1})
=> \left( { P }_{ 2 }{ P }_{ 1 } \right) \times 1=1\times 8.314\times (3727)
=> \left( { P }_{ 2 }{ P }_{ 1 } \right) = 83.14 Pa