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

An ideal gas has initial volume V and pressure P. In doubling its volume the minimum work done will be in the following process(of given processes)

  1. Isobaric process

  2. Isothermal process

  3. Adiabatic process

  4. None of the above.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Work done is the area under the P-V curve. For a given expansion, the adiabatic curve is the steepest, resulting in the smallest area under the curve compared to isothermal or isobaric processes.

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

Two difference gases of molecular masses $M _1$ and $M _2$ are at the same temperature. What is the ratio of their mean square speeds?

  1. $\dfrac{M _1}{M _2}$
  2. $\dfrac{M _2}{M _1}$
  3. $\sqrt {\dfrac{M _1}{M _2}}$
  4. $\sqrt {\dfrac{M _2}{M _1}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mean squared speed $=\cfrac{3RT}{M}$

$\cfrac{V _1}{V _2}=\cfrac{M _1}{M _2}$

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

A diatomic gas which has initial volume of $10$ litre is isothermally compressed to $1/15^{th}$ of its original volume where initial pressure is $10^5$ Pascal. If temperature is $27^o$C then find the work done by gas.

  1. $-2.70\times 10^3$J
  2. $2.70\times 10^3$J
  3. $-1.35\times 10^3$J
  4. $1.35\times 10^3$J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$w=nRT ln\left(\dfrac{v _2}{v _1}\right)$
$w=P _0V _0ln\left(\dfrac{v _2}{v _1}\right)$
$w=10^5\times 10\times 10^{-3}ln\left(\dfrac{1}{15}\right)$
$w=-2.70\times 10^3J$.

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 undergoes an isothermal change at temperature T so that its volume V is doubled. R is the molar gas constant. Work done by the gas during this change is :

  1. RT $\ln 4$
  2. RT $\ln 3$
  3. RT $ \ln 2$
  4. RT $ \ln 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Under isothermal process work is given by the relation $W = RT \ln (\dfrac{V _{f}}{V _{i}})$
Therefore work will be $W = RT \ln(2)$

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 $\left (C _{P} = \dfrac {7R}{2}\right )$ at pressure $P _{A}$ and temperature $T _{A}$ is isothermally expanded to twice the initial volume. The gas is then compressed at constant pressure to its original volume. Finally the gas is heated at constant volume to its original pressure $P _{A}$.
Calculate the net work done by the gas and the net heat supplied to the gas during the complete process.

  1. $0.579\ RT _{A}, \triangle Q = 0.579\ RT _{A}$.
  2. $79\ RT _{A}, \triangle Q = 0.679\ RT _{A}$.
  3. $0.9\ RT _{A}, \triangle Q = 0.779\ RT _{A}$.
  4. $0.7\ RT _{A}, \triangle Q = 0.979\ RT _{A}$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The process involves isothermal expansion, isobaric compression, and isochoric heating. Calculating work for each step and summing them yields the net work; for a cycle, the heat supplied equals the work done if the internal energy change is zero, but here we sum the heat for each leg.

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 u ndergoes a process:
$P = \dfrac{P _0}{1+(V _0/ V)^2}$.
Here $P _0$ and $V _0$ are constants. change in temperature of the gas when volume is changed from $V=V _0$ to $V = 2V _0$ is: 

  1. $-\dfrac{2P _0V _0}{5R}$
  2. $\dfrac{11P _0V _0}{10R}$
  3. $-\dfrac{5P _0V _0}{4R}$
  4. $P _0V _0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the ideal gas law PV = nRT, we substitute P = P0 / (1 + (V0/V)^2). Then T = PV/nR = (P0 * V) / (nR * (1 + (V0/V)^2)). Evaluating at V=V0 and V=2V0 allows calculating the change in temperature.

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

Let $Q$ and $W$ denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.

  1. $Q = 0$
  2. $W = 0$
  3. $Q \neq W$
  4. $Q = W$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In an isothermal process for an ideal gas, the change in internal energy (dU) is zero because temperature is constant. According to the first law of thermodynamics (dQ = dU + dW), dQ must equal dW.

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

Work done during isothermal expansion of one mole of an ideal gas $10$ atm to $1$ atm at $300\ K$ is

  1. $-4938.8\ J$
  2. $4938.8\ J$
  3. $-5744\ J$
  4. $6257.2\ J$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Work done in isothermal expansion is W = nRT ln(P1/P2). Using n=1, R=8.314 J/molK, T=300K, and ln(10/1) = 2.303, we get W = 1 * 8.314 * 300 * 2.303 = 5744 J. Since it is expansion, the gas does work, but the question asks for work done during expansion (often defined as positive by convention in physics).

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 monoatomic gas is at $360K$ and a pressure of $10 ^ { 5 } Pa.$ It is compressed at constant pressure until its volume is halved. Taking $R$ as $8.3 J{ mol } ^ { - 1 }{ K } ^ { - 1 }$ and the initial volume of the gas as $3.0 \times 10 ^ { - 2 } { m } ^ { 3 }$ , the work done on the gas is

  1. $-1500 J$
  2. $+1500 J$
  3. $-3000 J$
  4. $+3000 J$
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

An ideal gas is taken from state $A$ (pressure $P$, volume $V$) to state $B$ (pressure $\displaystyle\frac{P}{2}$, volume $2V$) along a straight line path in the pressure-volume diagram. Select the correct statements from the following.

  1. The work done by the gas in the process $A$ to $B$ exceeds the work done that would be done by it if the system were taken from $A$ to $B$ along an isotherm.
  2. In the temperature-volume diagram, the path $AB$ becomes a part of a parabola.
  3. In the pressure-temperature diagram, the path $AB$ becomes a part of hyperbola.
  4. In going from $A$ to $B$, the temperature $T$ of the gas first increases to a maximum value and then decreases.
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

The given process follows a linear P-V relation given by: $ \dfrac{p-P}{v-V}=\dfrac{P-P/2}{V-2V}=-P/2V$
$\Rightarrow p-P=\dfrac{-P}{2V}(v-V) \Rightarrow p=-\dfrac{Pv}{2V}+3P/2$        .......(i)


Here work done $\Delta W$=area under P-V diagram =$3PV/4=0.75PV$
For isotherm, pv=PV $\Rightarrow$ work done $\Delta W$=$PV\ln2=0.693PV$

Hence statement a is correct
For t-v diagram, replace p in (i) by $\dfrac{nRt}{v}$
Hence, $t=-\dfrac{Pv^2}{2nRV}+\dfrac{3Pv}{2nR}$

The above eqn. has t equalling a quadratic in v. 
Hence, t-v diagram is a parabola $\Rightarrow$ statement b is correct.

Clearly the above parabola is concave downwards , hence has a maxima.
The t has same initial and final value implying the maxima occurs in-between.

Hence statement d is correct.

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

A fixed mass of a gas is first heated isobarically to double the volume and then cooled isochorically to decrease the temperature back to the initial value. By what factor would the work done by the gas decreased, had the process been isothermal?

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

Let initial pressure and volume be P  and V respectively.
Then after isobaric expansion they are P and 2V. Here work done=$P\Delta V=P(2V-V)=PV$
To bring to initial temperature new pressure=$\dfrac{PV}{2V}=P/2$. 


So after isochoric process they are $P/2  \ and \  2V$ . Here, $\Delta V=0\Rightarrow$ work done=$P\Delta V=0$
Hence, total work done in the 2 successive processes=PV +0=PV
For isothermal expansion, work done=$PVln(\dfrac{V _f}{V _i})=PV\ln2$
Hence required factor=$\dfrac{PV \ln2}{PV}=ln2$

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

Work done in reversible isothermal process by an ideal gas is given by

  1. $2.303 \text { nRT log } \frac { V _ { 2 } } { V _ { 1 } }$
  2. $\frac { n R } { ( y - 1 ) } \left( T _ { 2 } - T _ { 1 } \right)$
  3. $2.303 \text { nRT log } \frac { V _ { 1 } } { V _ { 2 } }$

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

The work done in a reversible isothermal process for an ideal gas is derived from the integral of PdV, resulting in W = nRT ln(V2/V1). Using base-10 logarithms, this is 2.303 nRT log(V2/V1).

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

A gas expands from $1l$ to $3l$ at atmospheric pressure. The work done by the gas is about 

  1. $2\ J$
  2. $200\ J$
  3. $300\ J$
  4. $2 \times 10^{5}\ J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work done = P * deltaV. P = 1 atm = 1.013 * 10^5 Pa. deltaV = 3L - 1L = 2L = 2 * 10^-3 m^3. W = 1.013 * 10^5 * 2 * 10^-3 = 202.6 J, which is approximately 200 J.

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

4 atm pressure to attain 1 atm pressure by result of isothermal  expansion. The work done by the gas during expansion is nearly.

  1. 155 J

  2. 255 J

  3. 355 J

  4. 555 J
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$V=1L=0.001m^3$
$P=4atm$
$\dfrac{P _1}{P _2}=4$
From ideal gas,
$PV=nRT$
$T=\dfrac{PV}{nR}$
Work done during the isothermal expansion,
$W=nRTln(\dfrac{P _1}{P _2})$
$W=nR.\dfrac{PV}{nR}ln(4)$
$W=PVln(4)$
$W=4\times 10^5\times 0.001\times ln(4)$
$W=555J$
The correct option is D.