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 physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

During the process A -B of an ideal gas: 

  1. work done on the gas is zero

  2. density of the gas is constant

  3. slope of line AB from the T - axis is inversely proportional to the number of moles of the gas

  4. slope of line AB from the T - axis is directly proportional to the number of moles of the gas
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

A gas follow a general process as $PV RT + 3V$ for $1\ mole$ of gas. If it expands isobarically till temperature is doubled, then the work done by the gas is (initial temperature and pressure are $T _{0}$ and $P _{0}$ respectively).

  1. $\dfrac {P _{0}T _{0}R}{(2P _{0} - 3)}$
  2. $\dfrac {P _{0}T _{0}R}{(P _{0} - 3)}$
  3. $\dfrac {P _{0}T _{0}R}{(P _{0}V - 3)}$
  4. $\dfrac {3P _{0}V _{0}}{R}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given PV = RT + 3V. For isobaric expansion, P is constant. Differentiating: P dV = R dT. Work done W = P * deltaV = R * deltaT. Since temperature doubles, deltaT = T0. Thus W = R * T0. The provided option A seems to be a complex derivation involving the specific equation of state.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

The work done by 100 calorie of heat in isothermal expansion of ideal gas is 

  1. 418.4J

  2. 4.184J

  3. 41.84J

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

One calorie is equivalent to $4.184$ joules 

or,   $J=4.18 J/cal$.
Work done (W)=$ = 4.18 \times 500 = 2090\,J$
As $1$ calorie is equal to $4.184$ calorie then $100$ calorie is equal to $ = 4.184 \times 100 = 418.4$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

The pressure and volume of a given mass of gas at a given temperature are $ \mathrm{P}  $ and $ \mathrm{V}  $ respectively.Keeping temperature constant, the pressure is increased by 10$  \%  $ and then decreased by 10$  \%  $ .The volume how will be -

  1. less than $ \mathrm{V} $
  2. more than $ \mathrm{V} $
  3. equal to $ \mathrm{V} $
  4. less than $ V $ for diatomic and more than $ V $ for monoatomic
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Two bulbs of volume V and 4V contains gas at pressures of 5 atm,. 1 atm and at temperatures of 300K and 400K respectively. When these bulbs are joined by narrow tube keeping their temperature at their initial values. The pressure of the system is

  1. 1 atm

  2. 2 atm

  3. 2.5 atm

  4. 3 atm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the ideal gas law for the two bulbs: n1 = P1V1/RT1 and n2 = P2V2/RT2. Total moles n = n1 + n2. Final pressure P = nRT_final / V_total. Since temperatures are kept constant, we use the weighted average pressure approach.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

One mole of an ideal gas at $300K$ is expanded isothermally from an initial volume of $1litre$ to $10litres$. The $\Delta E$ for this process is $(R=2cal.mol-1K-1)$

  1. $1381.1cal$

  2. zero
  3. $163.7cal$
  4. $9lit.atm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an ideal gas, internal energy E is a function of temperature only (E = f(T)). In an isothermal process, deltaT = 0, therefore deltaE = 0.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

An Ideal gas undergoes an isobaric process. If its heat capacity is $C _v$ at constant volume and number of mole $n$. then the ratio of work done by gas to heat given to gas when temperature of gas changes by $\Delta T$ is:

  1. $\left(\dfrac{nR}{c _v + R}\right)$
  2. $\left(\dfrac{R}{c _v + R}\right)$
  3. $\left(\dfrac{nR}{c _v - R}\right)$
  4. $\left(\dfrac{R}{c _v - R}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{f}{2} R = \dfrac{C _v}{n}$
$W = nR \Delta T$
$\Delta Q = \left(\dfrac{f}{2} + 1\right) nR \, \Delta T$
$\dfrac{W}{\Delta Q} = \left(\dfrac{2}{f + 2} \right) = \dfrac{2}{\dfrac{2C _v}{nR} + 2} = \left(\dfrac{nR}{C _v + R} \right)$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Three moles of an ideal gas kept at a constant temperature at $300 K$ are compressed from a volume of $4 L$ to $1 L$. The work done in the process is

  1. $-10368 J$
  2. $-110368 J$
  3. $12000 J$
  4. $120368 J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work done in an isothermal process is given by 
$\displaystyle{W = 2.3026nRT\log _{10}\frac{V _2}{V _1}}$
Here, $n = 3, R = 8.31 J/mol^oC$
$T = 300 K$, $V _1= 4 L$, $V _2 = 1 L$
Hence, $\displaystyle{W = 2.3026 \times 3 \times 8.31 \times 300 \times log _{10}\frac{1}{4}}$
= $17221.15 (-2\log _{10} 2$)
= $-17221.15 \times 2 \times 0.3010 = -10368J$ 

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

For an isothermal expansion of an ideal gas, mark wrong statement

  1. there is no change in the temperature of the gas

  2. there is no change in the internal energy of the gas

  3. the work done by the gas is equal to the heat supplied to the gas

  4. the work done by the gas is equal to the change in its internal energy
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$In\quad isothermal\quad expansion.\ The\quad temperature\quad is\quad constant,\ so\quad the\quad change\quad in\quad temperature,\quad \triangle T\quad =\quad 0,\ as\quad \triangle v\quad =\quad \frac { 3 }{ 2 } \quad R& T\ and\quad \triangle T\quad =\quad 0\ \therefore \quad \quad \triangle V\quad =\quad 0\ \ \quad \quad \quad \quad \quad \quad \quad \triangle V\quad =\quad Q-W\ \quad \quad \quad \quad \quad \quad \quad \quad 0\quad =\quad Q-W\ \quad \quad \quad \quad \quad \quad \quad \quad Q\quad =\quad W\ Work\quad done\quad in\quad isothermal\quad process\quad =\quad nRln\frac { { V } _{ 2 } }{ { V } _{ 1 } } \ \quad \quad \quad nRln\frac { { V } _{ 2 } }{ { V } _{ 1 } } \quad \neq \quad 0\quad \ \quad \therefore \quad \triangle W\quad \neq \quad \triangle V\ Answer\quad :\quad D$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

The cyclic process from X to Y is an isothermal process.
If the pressure of the gas at X is 4.0 kPa, and the volume is 6.0 cubic meters, and if the pressure at Y is 8.0 kPa, what is the volume of the gas at Y?

  1. 12.0 cubic meters

  2. 16.0 cubic meters

  3. 3.0 cubic meters

  4. 4.0 cubic meters

  5. 2.0 cubic meters
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given :    $P _x = 4.0$ kPa          $V _x = 6.0$  $m^3$                         $P _y = 8.0$ kPa 
X to Y is an isothermal process, thus temperature remains constant   i.e     $T = constant$  or   $PV = constant$
$\implies$     $P _yV _y = P _xV _x$
OR       $8.0 \times V _y  = 4.0 \times 6.0$                     $\implies V _y = 3.0$  $m^3$
Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

The work done y a gas is an isothermal change where 1 refers to initial state and 2 refers to final state is

  1. $\mu R({T _2} - {T _1})\ln \left( {\frac{{{V _2}}}{{{V _1}}}} \right)$
  2. $\mu R{T _1}\ln \left( {\frac{{{V _2}}}{{{V _1}}}} \right)$
  3. $\mu R{T _2}\ln \left( {\frac{{{V _1}}}{{{V _2}}}} \right)$
  4. $\mu R\left( {\frac{{{T _1} + {T _2}}}{2}} \right)[\ln {V _2} - \ln {V _1}]$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$w=-\int PdV   PV=nRT$
$T\rightarrow const$
$=-\int _{v _1}^{v _2}\frac {nRT}{V}dV$
$=\mu RT _1ln(\frac {V _2}{V _1})$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

An ideal gas system undergoes an isothermal process, then the work done during the process is:

  1. $nRT ln\dfrac { { V } _{ 2 }}{ { V } _{ 1 } }$
  2. $nRT ln\dfrac { { V } _{ 1 }}{ { V } _{ 2 }}$
  3. $2nRT ln\dfrac { { V } _{ 2 }}{ { V } _{ 1 }}$
  4. $2nRT ln\dfrac { { V } _{ 1 }}{ { V } _{ 2 }}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let an ideal gas goes isothermally from its initial state $(P _1V _1)$ to the final state $(P _2V _2)$. Then the work done,

$W=\int _{V _1}^{V _2}P\,dV$

We have 

$PV=nRT$

or

$P=\dfrac{nRT}{V}$

Then,

$W=\int _{V _1}^{V _2} \dfrac{nRT}{V} dV=nRT\int _{V _1}^{V _2}\dfrac 1V dV$

$=nRT \,ln(V _2)-ln(V _1)=nRT\,ln\dfrac{V _2}{V _1}$