Physics

Thermodynamics and Gas Laws

616 Questions

Thermodynamics and gas laws questions test the understanding of ideal gas behavior, work done during thermodynamic processes, and specific heat ratios. Key areas include isothermal, adiabatic, and isobaric expansions along with real gas deviations. These mathematical physics concepts are standard in engineering and general science competitive exams.

Ideal gas equationIsothermal and adiabatic processesThermodynamic workGas kinetic theoryReal gas behavior

Thermodynamics and Gas Laws Questions

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

A monatomic ideal gas expands at constant pressure, with heat Q supplied. The fraction of Q which goes as work done by gas is

  1. 1
  2. $\displaystyle{\dfrac{2}{3}}$
  3. $\displaystyle{\dfrac{3}{5}}$
  4. $\displaystyle{\dfrac{2}{5}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Q = nC _p \Delta T$ and $W = P\Delta V = nR\Delta T$
monatomic gas, $\displaystyle{C _p = \dfrac{5R}{2}}$.
$\Rightarrow$$\displaystyle{\dfrac{W}{Q} = \dfrac{2}{5}}$

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

Two moles of ideal helium gas are in a rubber balloon at $30^{o}C$. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to $35^{o}C$. The amount of heat required in raising the temperature is nearly $($take $R=8.31 J/ mo 1.K)$

  1. $62 J$
  2. $104 J$
  3. $124 J$
  4. $208 J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For isobaric process.
$ \Delta Q= n C _{p} \Delta T$
$=2 \times \dfrac{5}{2} R \times (35-30)$
$= 208 \ J$

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

The temperature of $5\ moles$ of a gas which was held at constant volume was changed from $100^{o}C$ to $120^{o}C$. The change in the internal energy of the gas was found to be $80\ J$, the total heat capacity of the gas at constant volume will be equal to

  1. $8\ J/K$
  2. $0.8\ J/K$
  3. $4.0\ J/K$
  4. $0.4\ J/K$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$dU = nC _v dT$ or, $ 80 = 5 \times C _v(120 - 100)$
$C _v = 4.0\ J/K$

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 at 30 K but decreases to 1.4 at 300 K as more degrees of freedom become active. During this rise in temperature

  1. $C _p$ remains constant but $C _v$ increases
  2. $C _p$ decreases by $C _v$ increases
  3. both $C _p$ and $C _v$ decreases by the same amount
  4. both $C _p$ and $C _v$ increases by the same amount
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

he value of the ratio $\dfrac{Cp}{Cv} $for hydrogen is 1.67 at 30 K but decreases to 1.4 at 300 K as more degrees of freedom become active. During this rise in temperature both $Cp$ and $Cv$ increases by the same amount
 For an ideal gas, $C _p = C _v + R$. If it is a molecular gas, increasing temperature enables vibrational degrees of freedom, so that $C _v$ increases. Hence $\dfrac{C _p}{C _v} = 1 +\dfrac{ R}{C _v}$ decreases.

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

If $ {C} _{P}$ and $ {C} _{V}$ denote the specific heats (per unit mass) of an ideal gas of molecular weight M then which of the following relations is true ?
(R is the molar gas constant)

  1. ${C} _{P}$ - ${C} _{V} = R$
  2. ${C} _{P}$ - ${C} _{V} = R / M$
  3. ${C} _{P}$ - ${C} _{V} = MR$
  4. ${C} _{P}$ - ${C} _{V}$ = $R /{M}^{2} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $Cu$ and $Cp$ be molar specific heats of the ideal gas at a 


constant volume and constant pressure, respectively, then

$C _p=M _{c _p}$ and $C _v=M _{c _v}$

Where $C _p$ and $C _v$ are specific heat (per unit mass)

if $C _p$ and $C _v$ are specific heat (for unit mass) of an ideal gas of molecular weight $M$

then specific heat (At constant P) for $M=MC _p$ and 

then specific heat (At constant V) for $M=MC _v$ 

then, $M _{C _p}-M _{C _v}=R$

$\boxed{C _p-C _v=R/M}$

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

If heat energy $\Delta $ is supplied to an ideal diatomic gas and the increase in internal energy is $\Delta U$, the ratio of $\Delta U:\Delta Q$ is

  1. $7:5$
  2. $5:7$
  3. $5/2 :7/2$
  4. $3:2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a diatomic gas, the specific heat at constant pressure $C _p=\frac{7}{2}R$ and the specific heat at constant volume $C _v=\dfrac{5}{2}R$

Thus, $\Delta U=nC _v\Delta T=\dfrac{5}{2}nR\Delta T$ and 
$\Delta Q=nC _p\Delta T=\dfrac{7}{2}nR\Delta T$
Hence, $\Delta U:\Delta Q=5/2:7/2$

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

$310 J$ of heat is required to raise the temperature of $2$ moles of an ideal gas at constant pressure from $25^0C$ to $35^0C$. The amount of heat energy required to raise the temperature of the gas through the same range at constant volume is

  1. $452J$
  2. $276J$
  3. $144J$
  4. $384J$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Heat = moles(no.) $\times C _P \times \triangle T$
$\Rightarrow 310=2\times { C } _{ P }\times 10\quad \quad [35-25=10]\\ \Rightarrow { C } _{ P }=15.5J/molK\\ $
$\therefore { C } _{ P }-{ C } _{ V }=R\\ \Rightarrow { C } _{ V }={ C } _{ P }-R=15.5-8.314\\ \Rightarrow { C } _{ V }=7.186J/molK\\ $
$Q=n{ C } _{ V }\triangle T\\ =2\times 7.186\times 10\\ =143.72J\approx 144J$
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

$C _p$ and $C _v$ are specific heats at constant pressure and constant volume respectively. It is observed that
$C _p-C _v=a$ for hydrogen gas
$C _p-C _v=b$ for nitrogen gas
The correct relation between a and b is :

  1. $a=28 b$
  2. $a=\dfrac{1}{14}b$
  3. $a=b$
  4. $a=14b$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
For any gas, $C _p-C _v=R$, which is the gas constant.
Hence be it any gas, hydrogen or nitrogen, its value is same.
Here, for ideal gas, $C _p – C _v = R/M$, where $M$ is the mass of one mole of gas.
Mass of one mole of hydrogen  $M = 2$ g and that of nitrogen  $M = 28$ g 
$\therefore$ $a =C _p -C _v= R/2$  (for hydrogen) 
And $b =C _p - C _v = R/28$  (for nitrogen)
  $\implies  a = 14b$
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

When a heat of Q is supplied to one mole of a monatomic gas $\left ( \gamma =5/3 \right )$, the molar heat capacity of the gas at constant volume is

  1. $ \dfrac{3R}{4}$
  2. $ \dfrac{5R}{4}$
  3. $ \dfrac{7R}{4}$
  4. $\dfrac{3R}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $\gamma =5/3$

i.e $\displaystyle \dfrac {C _p}{C _v}=\dfrac {5}{3}$
and $C _p-C _v=R$

$\therefore \displaystyle \dfrac {C _v+R}{C _v}=\dfrac {5}{3}$
$\displaystyle 1+\dfrac {R}{C _v}=\dfrac {5}{3}$

$C _v= \dfrac{3R}{2}$
Option D.

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

The molar specific heat of helium at constant volume is $3\ cal/mol^{o}C$ . Heat energy required to raise the temperature of 1gm helium gas by $1^{o}C$ at constant pressure is :

  1. 1.2 cal

  2. 1.25 cal

  3. 3 cal

  4. 4 cal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Heat added for a constant pressure process is,


$dQ=dU+dW$

$nC _p\Delta T=nC _v\Delta T+nR\Delta T$

Given 1 gm of Helium, number of moles$= 1/4 =0.25$

$R=2\ cal/mol-K$

$dQ=0.25[3(1)+2(1)]=0.25(5)=1.25\ cal$

Option B.

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

When 5 moles of gas is heated from $100^{o}C$ to $120^{o}C$ at constant volume, the change in internal energy is 200 J. The specific heat capacity of the gas is

  1. $5\space Jmol^{-1}K^{-1}$
  2. $4\space Jmol^{-1}K^{-1}$
  3. $2\space Jmol^{-1}K^{-1}$
  4. $1\space Jmol^{-1}K^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Change in internal energy is given by
$dU=nC _v\Delta T$
$200=5(C _v)20$
$C _v=2J/mol.K$
Option C.

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

$n _{1}$ and $n _{2}$ moles of two ideal gases of the thermodynamics constant $\gamma _{1}$ and $\gamma _{2}$ respectively are mixed. $C _{p}/ C _{v}$ for the mixture is

  1. $\dfrac {\gamma _{1} + \gamma _{2}}{2}$
  2. $\dfrac {n _{1}\gamma _{1} + n _{2}\gamma _{2}}{n _{1} + n _{2}}$
  3. $\dfrac {n _{1}\gamma _{2} + n _{2}\gamma _{1}}{n _{1} + n _{2}}$
  4. $\dfrac {n _{1}\gamma _{1}(\gamma _{2} + 1) + n _{2}\gamma _{2}(\gamma _{1} - 1)}{n _{1}(\gamma _{1} - 1) + n _{2}(\gamma _{1} - 1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

1g of $H _{2}$ gas is heated by $1^{o}C$ at constant pressure. The amount of heat spent in expansion of gas is

  1. $\dfrac{4.155}{4.18}cal$
  2. $\dfrac{4.7}{2.1}cal$
  3. $\dfrac{6.8}{2.2}cal$
  4. $\dfrac{1.26}{1.7}cal$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The amount of heat spent in the process will be,
Q = $ nR \Delta T $
Q = $ \dfrac{1}{2} \times 8.314 \times 1 $ J
The same value in calorie will be, Q = $ \dfrac{4.155}{4.185} $ cal

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

The volume of $1\ kg$ of hydrogen gas at $N.T.P$ is $11.2\ m^{3}$. Specific heat of hydrogen at constant volume is $10046J\ kg^{-1}K^{-1}$. Find the specific heat at constant pressure.

  1. $13.8\ kJ/kg-K$
  2. $14.2\ kJ/kg-K$
  3. $16.4\ kJ/kg-K$
  4. $18.3\ kJ/kg-K$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Mass of hydrogen $m=1\,kg$

Volume of hydrogen $V=11.2\,{{m}^{3}}$

Specific heat of hydrogen at constant volume ${{C} _{V}}=10046\,JK{{g}^{-1}}{{k}^{-1}}$

 We know that,

N.T.P condition as follows

  $ P=1.01\times {{10}^{5}}\,N/{{m}^{2}} $

 $ T={{25}^{0}}C=298\,K $

 Applying gas equation

 $PV=nRT$

 Putting the values in the above equation

 $ PV=nRT $

$ 1.01\times {{10}^{5}}\times 11.2=1\times R\times 298 $

$ R=3795.97\,J/kgk $

 Now according to the Mayer's law

 ${{C} _{P}}-{{C} _{V}}=R$

 Putting the values in the above equation

 $ {{C} _{P}}-{{C} _{V}}=R $

 $ {{C} _{P}}=3795.97+10046 $

$ {{C} _{P}}=13841.97\,JK{{g}^{-1}}{{k}^{-1}} $

Hence, the specific heat of hydrogen at constant pressure is $13841.97\ J/Kg-k$

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

Molar heat capacity of an ideal gas whose molar heat capacity at constant is $C _v$ for process $P=2e^{2v}$( where P is pressure of gas and V is volume of gas)

  1. $C _v + \dfrac{R}{1+2V}$
  2. $C _v + \dfrac{R}{2V}$
  3. $C _v + \dfrac{R}{V}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} By\, u\sin  g\, \, first\, law\, of\, Ther{ { moodynamics } }:- \ dQ=dw+dU \ and,\, also\,  \ dQ=nCdT \ dw=Pdv \ dU=n{ C _{ v } }dT \ Now,\, substituting\, them\, in\, the\, first\, law\, we\, get \ \Rightarrow nCdT=PdV+n{ C _{ n } }dT \ C=\frac { { PdV } }{ { ndT } } +{ C _{ v } } \ To\, find\, \, \frac { { PdV } }{ { ndT } } \, we\, will\, use\, the\, ideal\, gas\, equation \ PV=nRT \ 2V{ e^{ 2V } }=nRT\, \, \, \, \, \, \left[ { \, { { Re } }place\, \, P=2{ e^{ 2v } } } \right]  \ Differentiating\, both\, sides\, with\, respect\, to\, T \ 2\left( { { e^{ 2V } }+2V{ e^{ 2V } } } \right) \frac { { dV } }{ { dT } } =nR \ Now,\, from\, this\, we\, have \ \frac { { PdV } }{ { ndT } } =\frac { R }{ { 1+2v } }  \ So,\, we\, get \ C={ C _{ v } }+\frac { R }{ { 1+2v } }  \ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$