Thermodynamics JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Thermodynamics, free to read — no sign-in needed. The full chapter has 49 questions; sign in to attempt the remaining 44 in the exam simulator.

1. Q1JEE Advanced Adv 2023 (Paper 1)
   A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas $\left(\gamma =\dfrac{5}{3}\right)$ and one mole of an ideal diatomic gas $\left(\gamma =\dfrac{7}{5}\right)$. Here, $\gamma$ is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of $66$ Joule when heated at constant pressure. The change in its internal energy is _____ Joule.
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   Answer: 121

   Change in internal energy for the mixture can be written as,$∆U={n}_{1}{\left({C}_{v}\right)}_{1}∆T+{n}_{2}{\left({C}_{v}\right)}_{2}∆T$ $=\left({n}_{1}{\left({C}_{v}\right)}_{1}+{n}_{2}{\left({C}_{v}\right)}_{2}\right)∆T...\left(i\right)$ For isobaric process, work done $=P∆V$ $=\left({n}_{1}+{n}_{2}\right)R∆T...\left(ii\right)$ Divide $\left(i\right)$ by $\left(ii\right)$, we get $\dfrac{∆U}{W}=\dfrac{\left({n}_{1}{\left({C}_{v}\right)}_{1}+{n}_{2}{\left({C}_{v}\right)}_{2}\right)∆T}{\left({n}_{1}+{n}_{2}\right)R∆T}$ $\Rightarrow ∆U=\dfrac{W}{R}\left(\dfrac{\left({n}_{1}{\left({C}_{v}\right)}_{1}+{n}_{2}{\left({C}_{v}\right)}_{2}\right)}{{n}_{1}+{n}_{2}}\right)$ $=\dfrac{66}{R}\dfrac{\left[2\times \dfrac{3R}{2}+1\times \dfrac{5R}{2}\right]}{2+1}$ $=121J$
2. Q2JEE Advanced Adv 2020 (Paper 2)
   A spherical bubble inside water has radius $R$. Take the pressure inside the bubble and the water pressure to be ${p}_{0}.$ The bubble now gets compressed radially in an adiabatic manner so that its radius becomes $(R-a)$. For $a≪R$ the magnitude of the work done in the process is given by $\left(4\pi {P}_{0}R{a}^{2}\right)X,$ where $X$ is a constant and $\gamma ={C}_{p}/{C}_{V}=41/30$. The value of $X$ is______
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   Answer: 2.05

   $\begin{aligned} &\mathrm{W}=(\Delta \mathrm{P})_{\text {avg }} \times 4 \pi \mathrm{R}^2 \mathrm{a} \\ &=\left|\frac{\mathrm{dP}}{2} \cdot 4 \pi \mathrm{R}^2 \mathrm{a}\right| \end{aligned}$ (for small change $(\Delta \mathrm{P})_{\text {avg }}<\mathrm{P}>$ arithmetic mean) $=\mathrm{PV} \text { gamma }=\mathrm{c} \Rightarrow \mathrm{dP}=-\gamma \frac{\mathrm{P}}{\mathrm{V}} \mathrm{dV}=-\frac{\gamma \mathrm{P}_0}{\mathrm{~V}} 4 \pi \mathrm{R}^2 \mathrm{a}$ $=\frac{\gamma \mathrm{P}_0}{2 \mathrm{~V}} \times 4 \pi \mathrm{R}^2 \mathrm{a} \times 4 \pi \mathrm{R}^2 \mathrm{a}$ $=\frac{\gamma \mathrm{P}_0}{2 \times 4 \pi \mathrm{R}^3} 4 \pi \mathrm{R}^2 \mathrm{a} \times 4 \pi \mathrm{R}^2 \mathrm{a}$ $=\left(4 \mathrm{pRP} \times \mathrm{a}^2\right) \frac{3 \gamma}{2}$ $\therefore \mathrm{x}=2.05$
3. Q3JEE Advanced Adv 2019 (Paper 2)
   A mixture of ideal gas containing $5$ moles of monatomic gas and $1$ mole of rigid diatomic gas is initially at pressure ${P}_{0},$ volume ${V}_{0}$ and temperature ${T}_{0}.$ If the gas mixture is adiabatically compressed to a volume $\dfrac{{V}_{0}}{4}$, then the correct statement(s) is/are, (Given ${2}^{1.2}=2.3\text{;}{2}^{3.2}=9.2\text{;}R$ is gas constant)
   1. A.The final pressure of the gas mixture after compression is in between $9{P}_{0}$ and $10{P}_{0}$.
   2. B.The average kinetic energy of the gas mixture after compression is in between $18R{T}_{0}$ and $19R{T}_{0}$.
   3. C.The work $\left|W\right|$ done during the process is $13R{T}_{0}$.
   4. D.Adiabatic constant of the gas mixture is $1.6$.
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   Answer: A,C,D

   $\begin{matrix}\text{Here,}{n}_{1}=5, & {C}_{{P}_{1}}=\dfrac{5}{2}R, & {C}_{{V}_{1}}=\dfrac{3}{2}R \\ {n}_{2}=1, & {C}_{{P}_{2}}=\dfrac{7}{2}R, & {C}_{{V}_{2}}=\dfrac{5}{2}R\end{matrix} {\gamma }_{mix}=\dfrac{{n}_{1}{C}_{{P}_{1}}+{n}_{2}{C}_{{P}_{2}}}{{n}_{1}{C}_{{V}_{1}}+{n}_{2}{C}_{{V}_{2}}}=\dfrac{8}{5}=1.6$ Work done, $W=\dfrac{{P}_{1}{V}_{1}-{P}_{2}{V}_{2}}{\gamma -1},\left(Inadiabaticprocess\right)$ Now, for adiabatic compression ${P}_{1}{V}_{1}^{r}={P}_{2}{V}_{2}^{r}$ $\Rightarrow {P}_{0}{V}_{0}^{8/5}={P}_{2}{\left(\dfrac{{V}_{0}}{4}\right)}^{8/5}$ $\Rightarrow {P}_{2}=9.2{P}_{0}\Rightarrow \left(A\right)$ is correct. Now, $W=\dfrac{{P}_{0}{V}_{0}-9.2{P}_{0}\dfrac{{V}_{0}}{4}}{3/5}=-13R{T}_{0}\Rightarrow \left(C\right)$ is correct $∴\left|W\right|=13R{T}_{0}$ Again ${T}_{1}{V}_{1}^{\gamma -1}={T}_{2}{V}_{2}^{\gamma -1}$ ${T}_{2}={T}_{1}{\left(2\right)}^{6/5}=2.3{T}_{0}$ Average kinetic energy of gas mixture $=n{C}_{{V}_{mix}}{T}_{2}$ $=23R{T}_{0}\Rightarrow \left(B\right)$ is incorrect
4. Q4JEE Advanced Adv 2018 (Paper 2)
   One mole of an ideal monatomic gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is $100K$ and the universal gas constant $R=8.0Jmo{l}^{-1}{K}^{-1}$, then how much is the decrease in its internal energy (in $J$) ?
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   Answer: 900

   ${V}_{i}=V$ ${V}_{f}=8V$ For adiabatic process $\{\gamma =5/3\text{ for monoatomic gas}\}$ ${T}_{1}\cdot {V}_{1}^{\gamma -1}={T}_{2}\cdot {V}_{2}^{\gamma -1}$ $100{\left(V\right)}^{\dfrac{2}{3}}={T}_{2}{\left(8V\right)}^{\dfrac{2}{3}}$ ${T}_{2}=25K$ Loss in internal energy $∆U=n{C}_{v}∆T=1\left(\dfrac{fR}{2}\right)\left[100-25\right]=12\times 75=900J$
5. Q5JEE Advanced Adv 2012 (Paper 2)
   Two moles of ideal helium gas are in a rubber balloon at $30^{\circ} \mathrm{C}$. 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 $35^{\circ} \mathrm{C}$. The amount of heat required in raising the temperature is nearly (take $R=8.31 \mathrm{~J} / \mathrm{mol} . \mathrm{K}$ )
   1. A.$62 \mathrm{~J}$
   2. B.$104 \mathrm{~J}$
   3. C.$124 \mathrm{~J}$
   4. D.$208 \mathrm{~J}$
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   Answer: (D)

   The heat is supplied at constant pressure. i.e., the process is isobaric $\begin{array}{l} \therefore Q=n C_{p} \Delta t \\ =2\left[\frac{5}{2} R\right] \times \Delta t=2 \times \frac{5}{2} \times 8.31 \times 5=208 \mathrm{~J} \\ \left(\because C_{p}=\frac{5}{2} R \text { for mono-atomic gas }\right) \end{array}$