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

The molar specific heat capacity varies as $C=C _v + \beta V$ ($\beta$ is a constant). Then the equation of the process for an ideal gas is given as

  1. $T^{\frac{\beta}{RV} }= constant$
  2. $V^{\frac{\beta T}{R}}=constant$
  3. $T^{-\frac{R}{\beta V}}=constant$
  4. $V^{\frac{R}{\beta T}}=constant$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the first law dQ = dU + dW, where dQ = C dT, dU = Cv dT, and dW = P dV. Substituting C = Cv + beta*V, we get (Cv + beta*V) dT = Cv dT + P dV. This simplifies to beta*V dT = P dV. Using P = RT/V, we get beta*V dT = (RT/V) dV, which leads to dT/T = (R/beta) * (dV/V^2). Integrating this leads to the process equation.

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

The temperature of 5  mole of a gas which was held at constant volume was change from ${ 100 }^{ 0 }$ C to $120^{ 0 }$ C the change in internal energy was found to be 80 joules 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

Total heat capacity at constant volume (Cv_total) is defined as Q / delta T. Given Q = 80 J and delta T = 20 K, Cv_total = 80 / 20 = 4.0 J/K.

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

When $1\ mole$ of a monoatomic gas expands at constant pressure the ratio of the heat supplied that increases the internal energy of the gas and that used in expansion is

  1. $\dfrac{2}{3}$
  2. $\dfrac{3}{2}$
  3. $0$
  4. $\infty$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Heat supplied to one mole of gas in a constant pressure process is given by: $Q = C _{p}\Delta T$
Change in the internal energy of gas is given by:$\Delta U = C _{v}\Delta T$


The ratio of heat that goes into increasing the internal energy is:
$\dfrac{\Delta U}{Q} = \dfrac{C _{v}}{C _{p}} = \dfrac{1}{\gamma}$

For a mono atomic gas $\gamma = \dfrac{5}{3}$
So, $\dfrac{3}{5}$ ratio of heat goes into increasing the internal energy, and the rest goes into expansion work = $\dfrac{2}{5}$ of heat supplied

Hence, the ratio of heat supplied to increase internal energy by heat supplied to do expansion is $ = \dfrac{3}{2}$

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

One mole of helium is heated at $0^o$C and constant pressure. How much heat is required to increase its volume threefold?

  1. $3820\ cal$
  2. $382\ cal$
  3. $38.2\ cal$
  4. $3.28\ cal$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As it is a constant pressure process, using Charles law we get $\displaystyle\frac{V}{T}=constant$. Thus for a threefold increase in volume we get threefold increase in temperature. Thus we get the final temperature as $3(273)=819 K.$ Thus $\Delta T=819-273=546 K$.
Now as helium is a diatomic molecule, its degree of freedom f is 5. Ths we get $C _p$ for it as $(1+\displaystyle\frac{f}{2})R=(1+\frac{5}{2})R=\frac{7}{2}R$.
Thus heat transferred will be given as $\Delta Q=nC _p\Delta T$
or
$\Delta Q=1(\displaystyle\frac{7}{2})(1.987)(546)=3820\  cal$

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

When an ideal diatomic gas is heated at constant pressure then what fraction of heat given is used to increase internal energy of gas ? 

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

For a diatomic gas we have the degree of freedom as 5. 

Thus heat given at constant pressure is given as $nC _p\Delta T=n(1+\displaystyle\dfrac{5}{2})R\Delta T=n\dfrac{7}{2}R\Delta T$. 
The heat given to change the internal energy is $nC _v\Delta T=n\displaystyle\dfrac{5}{2}R\Delta T$. 
The fraction of internal energy thus used is $\dfrac{5}{7}$ 

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

One mole of a monoatomic gas and one mole of a diatomic gas are mixed together. What is the molar specific heat at constant volume for the mixture ?

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

$C _v$ is given as $\displaystyle\dfrac{f}{2}R$. Here $f$ is the degree of freedom. For monoatomic gas $f=3$ and for diatomic gas $f=5$. 

Thus we get $C _v$ for the mixture as $\displaystyle\dfrac{n _1}{n _1+n _2}(\dfrac{3}{2}R+\dfrac{5}{2}R)=2R$. Here $n _1$ and $n _2$ both are 1.

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

Equal volumes of monoatomic and diatomic gases of same initial temperature and pressure are mixed. The ratio of the specific heats of the mixture ($C _p/C _v$) will be

  1. $1.53$
  2. $1.52$
  3. $1.5$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$C _p$ for a gas is given as $(1+\displaystyle\dfrac{f}{2})R$ and $C _v$ is given as $\displaystyle\dfrac{f}{2}R$. Here f is the degree of freedom. 


For monoatomic gas it is 3 and for a diatomic gas it is 5. 

Thus, we get $C _p$ for the mixture as $\displaystyle\dfrac{5}{2}R+\dfrac{7}{2}R=6R$ and $C _v$ is given as $\displaystyle\dfrac{3}{2}R+\dfrac{5}{2}R=4R$.

Thus, the ratio $\displaystyle\dfrac{C _p}{C _v}$ is given as $\displaystyle\dfrac{3}{2}=1.5$