Tag: heat and thermodynamics

Quizzes (20) Questions (162)

Questions Related to heat and thermodynamics

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A monoatomic gas expands at a constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the expansion is 

  1. $75\%, 25\%$

  2. $25\%, 75\%$

  3. $60\%, 40\%$

  4. $40\%, 60\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to the first law of thermodynamics,
Q = U + W
Now, at constant pressure, 
W = $ P \Delta V = nR \Delta T $
U = $ n {C} _{v} \Delta T $
For, a monoatomic gas, $ {C} _{v} = 1.5 R $
Thus, Q = $ 2.5 nR \Delta T $
Now, $ \dfrac{U}{Q} = \dfrac{1.5R}{2.5R}  = 60 \%$
Similarly, $ \dfrac{W}{Q} = \dfrac{R}{2.5 R} = 40 \%$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

If for hydrogen $C _p-C _v=m$ and for nitrogen $C _p-C _v=n$, where $C _p$ and $C _v$ refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between $m$ and $n$ is (molecular weight of hydrogen$=2$ and molecular weight of nitrogen$=14$)

  1. $n=14m$

  2. $n=7m$

  3. $m=7n$

  4. $m=14n$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For hydrogen, $C _P-C _V=\dfrac{1}{M _{H _2}}\dfrac{dQ}{dT}=m$

For nitrogen, $C _P-C _V=\dfrac{1}{M _{N _2}}\dfrac{dQ}{dT}=n$
$\implies \dfrac{m}{n}=\dfrac{M _{N _2}}{M _{H _2}}=\dfrac{14}{2}=7$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The average degree of freedom per molecule for a gas are $6$. The gas performs $25 J$ of work when it expands at a constant pressure. The heat absorbed by gas is 

  1. $75 \ J$

  2. $100 \ J$

  3. $150\ J$

  4. $125 \ J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a gas with 'n' degrees of freedom:
$\gamma = 1 + \dfrac{2}{n} = 1 + \dfrac{2}{6} = \dfrac{4}{3}$
$C _{p} = \dfrac{\gamma R}{\gamma - 1} = 4R$
$C _{v} = \dfrac{R}{\gamma - 1} = 3R$

Heat supplied for constant pressure process is $nC _{p}\Delta T$

Change in internal energy $nC _{v} \Delta T$
$\dfrac{\Delta U}{Q} = \dfrac{C _{v}}{C _{p}} = \dfrac{1}{\gamma} = \dfrac{3}{4}$

Hence $\dfrac{W}{Q} = 1 - \dfrac{\Delta U}{Q} = \dfrac{1}{4}$
$\dfrac{W}{Q}=1-\dfrac{3}{4}=\dfrac{1}{4}$
$\implies Q = 100J$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

What is the ratio of specific heats of constant pressure and constant volume for $NH _3$

  1. 1.33

  2. 1.44

  3. 1.28

  4. 1.67
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Ammonia (NH3) is a polyatomic gas. The ratio of specific heats (gamma) for polyatomic gases is typically lower than that of diatomic gases (1.4), and 1.28 is the standard value for NH3.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A reversible adiabatic path on a P- V diagram foran ideal gas passes through state A where P = 0.7$\times $ ${ 10 }^{ 2  }$ N/${ m }^{ -2 }$ and v=0.0049 $ { m }^{ 3  }$, The ratio of specific heat of the gas is 1.4 , The slop of patch at A is:

  1. $2.0 \times{ 10 }^{ 3\quad }{ Nm }^{ -5 }$

  2. $1.0 \times{ 10 }^{ 3\quad }{ Nm }^{ -8}$

  3. $-2.0\times{ 10 }^{ 7\quad }{ Nm }^{ -3 }$

  4. $-1.0\times{ 10 }^{ 3\quad }{ Nm }^{ -5 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The value of the ratio ${C} _{p}/{C} _{v}$ for hydrogen is $1.67$ a $30K$ but decreases to $1.4$ at $300K$ as more degrees of freedom become active. During this rise in temperature (assume H2 as ideal gas),

  1. ${C} _{p}$ remains constant but ${C} _{v}$ increases

  2. ${C} _{p}$ decreases but ${C} _{v}$ increases

  3. Both ${C} _{p}$ and ${C} _{v}$ decreases by the same amount

  4. Both ${C} _{p}$ and ${C} _{v}$ increase by the same amount
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A polyatomic gas with six degrees of freedom does $25\ J$ of work when it is expanded at constant pressure. The heat given to the gas is

  1. $100\ J$

  2. $150\ J$

  3. $200\ J$

  4. $250\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Degree of freedom, $f=6$
$\Rightarrow C _v=\dfrac{fR}{2}=3R$
$\Rightarrow C _p=C _v+R=4R$
Also work done $=\Delta W$=25J


Thus for isobaric process applying first law,
Heat given($\Delta Q$) $=$ internal energy change$(\Delta U)+\Delta W$
$\Rightarrow nC _p\Delta T=nC _v\Delta T+\Delta W$
$\Rightarrow 4nR\Delta T=3nR\Delta T +25J$
$\Rightarrow nR\Delta T=25J$
Hence, $ \Delta Q=4\times 25=100J$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A gas expands against a constant external pressure of  $2.00 atm, $ increasing its volume by $ 3.40 L.$   Simultaneously, the system absorbs  $400 J $ of heat from its surroundings. What is  $ \Delta E ,$  in joules, for this gas?

  1. $- 689$

  2. $-289$

  3. $+400$

  4. $+289$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the first law of thermodynamics, delta E = Q - W. Work done by the gas is P * delta V = 2.00 atm * 3.40 L = 6.80 L*atm. Converting to Joules (1 L*atm = 101.3 J), W = 6.80 * 101.3 = 688.84 J. Thus, delta E = 400 J - 688.84 J = -288.84 J, which rounds to -289 J.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

Consider a classroom that is roughly  $5 { m } \times 10  { m } \times 3  { m }.$  Initially   ${ t } = 20 ^ { \circ }  { C }$  and  $ { P } = 1$ atm. There are  $50$  people in an insulated class loosing energy to the room at the average rate of  $150$  watt per person. How long can they remain in class if the body temperature is  $37 ^ { \circ } \mathrm { C }$  and person feels uncomfortable above this temperature. Molar heat capacity of air  $= ( 7 / 2 ) R.$

  1. $4.34$ minutes

  2. $5.73$ minutes

  3. $6.86$ minutes

  4. $7.79$ minutes
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The room volume is 150 m^3. Using PV=nRT, calculate the number of moles of air. The total heat added by 50 people is 50 * 150 W = 7500 J/s. The heat required to raise the air temperature from 20 C to 37 C is Q = n * Cv * delta T. Solving for time t = Q / Power gives approximately 4.34 minutes.