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Thermodynamics

Description: Engineering Entrance Physics
Number of Questions: 17
Created by:
Tags: Engineering Entrance Physics Liquids and Gases Thermal Expansion Thermal Expansion of Solids Thermodynamics Heat Thermal Properties of Matter Ideal Gas Laws Ideal Gas Equation and Absolute Temperature
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A temperature difference of 5oC on Celsius scale corresponds to the following temperature in the Fahrenheit scale.

  1. 41

  2. 32.05

  3. 33

  4. 77

  5. 13

Show answer
Correct Option: A
Explanation:

(0C/100) = (0F – 32)/180 5/100  = (0F – 32)/180 9 = 0F – 320F = 41

In a temperature scale called Z, the boiling point of water is at 65oZ and the freezing point is at - 14oZ. Then the temperature T = - 98oZ corresponds on the Fahrenheit scale to

    • 159oF
    • 363oF
    • 223oF
    • 427oF
    • 72oF
Show answer
Correct Option: A
Explanation:

 Z = -98 0 {Z – (- 14)}/{65 – (- 14)} = (F - 32)/180 (- 98 +14)/(65 + 14) = (F - 32)/180 - 84/79 = (F - 32)/180 (- 84/79) x 180 = (F- 32) - 191 + 32 = F F =  - 1590F

A tap supplies water at 10oC and another tap at 100oC. How much hot water must be taken so that we get 20 kg of water at 35oC?

  1. 5.56 kg

  2. 8.18 kg

  3. 12.5 kg

  4. 45 kg

  5. 4.54 kg

Show answer
Correct Option: A
Explanation:

 Let mass of hot water be m kg. Mass of cold water = (20 - m)kg Heat taken by cold water               = (20 - m) x 1 x (35 – 10) Heat given by hot water              = m x 1 x (100 – 35) Law of mixture gives Heat given by hot water = Heat taken by cold water m x 1 x (100 – 35) = (20 - m) x 1 x (35 – 10) 65 m = (20 - m) x 25 65 m = 500 – 25 m 90 m = 500      m = 500/90         = 5.56 kg

The first excited state of hydrogen atom is 10.2 e V above the ground state. What temperature is needed to excite hydrogen atoms to first excited level?

  1. 7.88 x 104 K

  2. 3.94 x 104 K

  3. 5.91 x 104 K

  4. 23.65 x 104 K

  5. 17.73 x 104 K

Show answer
Correct Option: A
Explanation:

Where K = 1.38 x 10-23 J/mole. 0K K.E. per atom = (3/2) KT K.E. of the hydrogen atom = 10.2 eV ∴ 10.2 eV = 10.2 x (1.6 x 10-19) joule T = (2/3) x ( K.E. per atom/ K)   = (2/3) x 10.2 x (1.6 x 10-19)/1.38 x 10-23  = 7.88  x 104 K

At what temperature will the root mean square velocity of oxygen molecules be sufficient, so as to escape from the earth?

  1. 1.6 x 105 K

  2. 28.63 K

  3. 14.31 K

  4. 21.47 K

  5. 42.95 K

Show answer
Correct Option: A
Explanation:

 ∴ (3/2)KT = (1/2)mve2 Where ve = escape velocity of earth = 11.1 km/s K = 1.38 x 10-23 m = mass of 1 molecules of oxygen = 5.34 x 10-26 T  = (mve2/3K)     = 5.34 x 10-26(11.1 x 103)2/3 x 1.38 x 10-23

     = 1.6 x 105 K

A gas is filled in a container at pressure P0. If the mass of each molecule is halved and speeds are double, then the resulting pressure will be

  1. 2 Po

  2. 4 Po

  3. Po

  4. Po/2

  5. 8 Po

Show answer
Correct Option: A
Explanation:

Po = (1/3)(Nm/V) v2 rms (Here m = mass of each molecule, V = volume of the gas) Now , m’ = (1/2)m V’ rms = 2 v rms P’ = (1/3)(Nm/2V)4 v2 rms      = 2 x (1/3)(Nm/V) v2 rms = 2 Po

The density of carbon dioxide gas at 0oC and at a pressure of 1.0 x 105 Newton/metre2 is 1.98kg/m3. Find the root mean square velocity of its molecules at 0oC.

  1. 389 m/s

  2. 1.5 x 105 m/s

  3. 5.05 x 104 m/s

  4. 224 m/s

  5. 1.68 x 104 m/s

Show answer
Correct Option: A
Explanation:

= 389 m/s

A flask of 10-3 m3 volume contains 3 x 1022 molecules of oxygen at a certain temperature. The mass of one molecule of oxygen is 5.3 x 10-26 kg and rms velocity of its molecules at the same temperature is 400m/s. Calculate the pressure of the gas.

  1. 8.48 x 104 N/m2

  2. 25.44 x 104 N/m2

  3. 636 N/m2

  4. 212 N/m2

  5. 2.12 x 10-4 N/m2

Show answer
Correct Option: A
Explanation:

Pressure of the gas P = (1/3)(mN/V)v2 rms

Here  V =10-3m3 N = 3 x 1022 m = 5.3 x 10-26 kg v rms = 400 m/s ∴ P = (1/3){ 5.3 x 10-26 x 3 x 1022 (400)2}/10-3        = 8.48 x 104 N/m2

Calculate the kinetic energy of a gram molecule of argon at 127oC.

  1. 4986 J

  2. 9972 J

  3. 1662 J

  4. 6648 J

  5. 2216 J

Show answer
Correct Option: A
Explanation:

Temperature, T = 1270C                               = 273 + 127                               = 400 K R = 8.31 J/mol K Now K.E. per gram molecule of argon = (3/2) x 8.31 x 400 J             = 4986 J

5 g of ice at 0oC is dropped in a beaker containing 20 g of water at 40oC. What will be the final temperature?

  1. 16oC

  2. 26.66oC

    • 16oC
    • 26.66oC

  5. 35.29oC

Show answer
Correct Option: A
Explanation:

Let final temperature be θ. Heat taken by ice = m1L + m1c1∆θ1

                = 5 x 80 + 5 x 1(θ-0)

                 = 400 + 5θ

 Heat given by water at 400 C

                 = m2c2∆θ2

                 = 20 x 1 x (40-θ)

                = 800 -20θ

 Heat given = Heat taken

800 -20θ   =  400+ 5θ

 800-400   =   5θ + 20θ

        400      = 25θ

               θ     = 400/25

                      = 160 C

At what temperature does the average translational kinetic energy of a molecule in a gas equal to the kinetic energy of an electron accelerated from rest through a potential difference of 5 volt?

  1. 38.65 x 103 K

  2. 8.69 x 104 K

  3. 1.93 x 104 K

  4. 2.89 x 104 K

  5. 11.59 x 104 K

Show answer
Correct Option: A
Explanation:

 K.E. of the electron is 5 eV = 5 x 1.6 x 10-19 J But K.E. = (3/2)KT ∴ 5 x 1.6 x 10-19 = (3/2)(1.38 x 10-23) x T

T =  2 x 5 x 1.6 x 10-19/3 x 1.38 x 10-23

   = 38.65  x 103 K

If the pressure of a gas containing vessel increases by 0.5% when heated by 2oC, then the initial temperature must be

  1. 127oC

  2. 400oC

  3. 673oC

  4. 1073oC

  5. 527oC

Show answer
Correct Option: A
Explanation:

t1 = 400 - 273                   = 1270C

The pressure of a gas is P N/m2. The mean kinetic energy of one gram – mole at NTP ( in joule) will be

  1. 3.36 x 10-2 P

  2. 2.24 x 10-2 P

  3. 6.72 x 10-2 P

  4. 4.48 x 10-2 P

  5. 2.24 x 10-2 P

Show answer
Correct Option: A
Explanation:

K.E. = (f/2)RT        = (3/2)RT        = (3/2)PV        VNTP = 22.4 x 10-3 J K.E. = (3/2)(P) VNTP         = 33.6 x 10-3 P        = 3.36 x 10-2 P

At what temperature will the root mean square velocity of hydrogen be double of its value at S.T.P. , pressure remaining constant?

  1. 819oC

  2. 1092oC

  3. 1365oC

  4. 819oC

  5. 273oC

Show answer
Correct Option: A
Explanation:

 Let V1 be the r.m.s. velocity at S.T.P. and  V2 be the r.m.s. velocity at unknown temperature T2. Here, T1 = 273 K T2 =?  (V2 / V1 )= 2 V2 rms = 3KT/m We know, v2 ∝ T ∴ V12 / V22 = T1/T2

 T2 = T1 (V2 / V1 )2       = 273 x 22       = 273 x 4       = 1092 K T2 = 1092 - 273     =  8190C

The specific heat of argon at constant volume is 0.075 kcal/kg K. Calculate the atomic weight. [Given: R = 2cal/mol K]

  1. 40 gm/mole

  2. 53.33 gm/mole

  3. 17.77 gm/mole

  4. 0.225 gm/mole

  5. 0.1 gm/mole

Show answer
Correct Option: A
Explanation:

As argon is monatomic, its molar specific heat at constant volume will be Cv= (3/2)R = (3/2)2 = 3 cal/mol K Now as Cv = Mcv with Cv = 0.075 cal/gk              3 = M x 0.075          Mw = 3 / 0.075                =  40 gm/mole

A gram molecule of a gas is allowed to expand adiabatically at 327oC till the volume is four times the initial volume. Calculate the amount of heat absorbed. J = 4.2 x 107 ergs/cal, R = 8.31 x 107 ergs/gm mole oC and γ = 1.5

  1. 1187.14 cal

  2. 20941.2 x 1014 cal

  3. 593.57 cal

  4. 2374.28 cal

  5. 3561.42 cal

Show answer
Correct Option: A
Explanation:

 T2 = T1 (V1/ V2)γ -1 = 300K = 270C Hence work done, W = {R/(γ - 1)}( T1 – T2)      = {8.31 x 107/(1.5 - 1)} x 300      = 4986  x 107

Heat absorbed, H = W/J     = 4986 x 107/4.2 x 107    = 1187.14 cal

At the top of a mountain, a thermometer reads 7oC and barometer reads 70 cm of Hg. At the bottom of the mountain, they read 27oC and 76 cm of Hg. Compare the air at the top with that at the bottom.

  1. 0.9868

  2. 3.55

  3. 1.01

  4. 0.28

  5. 0.072

Show answer
Correct Option: A
Explanation:

 According to gas equation, PV =  µRT PV = (M/Mw)RT      [as µ = M/Mw] Or P/ρT = R/Mw    [as M/V = ρ] Now as M and R are same for top and bottom (P/ρT)T = (P/ρT)B  i.e., ρT /ρB  = (PT/PB) x (TB/TT) So, ρT/ρB = (70/76) x (300/280)                 = 75/76                =  0.9868  

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