Thermodynamics MCQs for NEET — Physics Questions with Answers

$U=\frac{n\text{fR}T}{2}$

⇒ ${\left(U\right)}_N=\frac{56\times {10}^3}{28}\times \frac{5}{2}R\times 300$

and ${\left(U\right)}_{Ar}=\frac{6\times {10}^{26}}{6\times {10}^{23}}\times \frac{3}{2}R\times 900\Rightarrow {\left(U\right)}_{Ar}<{\left(U\right)}_N$

A is compressed isothermally, hence

$P_{1}V=P_2\frac{V}{2}\Rightarrow P_2=2P_1$

and B is compressed adiabatically, hence

$P_1{V}^{\gamma }=P_2{\left(\frac{{\displaystyle V}}{{\displaystyle 2}}\right)}^{\gamma }\Rightarrow P_2={\left(2\right)}^{\gamma }{P}_1$

Since $\gamma >1$, hence $P_2\text{'}>P_2$ or $P_2<{P^{\text{'}}}_2$

In isothermal process $P_1{V}_1=P_2{V}_2$

or $PV=P_2\times 4V$

∴ $P_2=\frac{P}{4}$

In adiabatic process

$P_2{V_2}^{\gamma }=P_3{V_3}^{\gamma }$

⇒ $\frac{P}{4}\times {\left(4V\right)}^{1.5}=P_2{V}^{1.5}$

⇒ $P_3=2P$

Initially $\eta =\left(1-\frac{{\displaystyle T_2}}{{\displaystyle T_1}}\right)=\frac{W}{Q}=\frac{1}{6}$  ...(i)

Finally $\eta \text{'}=\left(1-\frac{{\displaystyle T_2\text{'}}}{{\displaystyle T_1}}\right)=\left(1-\frac{{\displaystyle (T_2-62)}}{{\displaystyle T_1}}\right)=1-\frac{T_2}{T_1}+\frac{62}{T_1}$

$=\eta +\frac{62}{T_1}$ ....(ii)

It is given that $\eta \text{'}=2\eta .$ Hence solving equation (i) and (ii)

⇒ $T_1=372K=99°C$ and $T_2=310K=37°C$

$P{V}^{5/3}=$ constant represents adiabatic equation.

So during the expansion of ideal gas internal energy of gas decreases and temperature falls.

P-V diagram of the gas is a straight line passing through origin. Hence $P\propto V$ or $P{V}^{-1}=$ constant

Molar heat capacity in the process $P{V}^x=\text{constant is }$

$C=\frac{R}{\gamma -1}+\frac{R}{1-x}$; Here $\gamma =1.4$ (For diatomic gas)

⇒ $C=\frac{R}{1.4-1}+\frac{R}{1+1}$ ⇒ $C=3R$

In both cylinders A and B the gases are diatomic (γ = 1.4). Piston A is free to move i.e. it is isobaric process. Piston B is fixed i.e. it is isochoric process. If same amount of heat ΔQ is given to both then

${\left(\Delta Q\right)}_{\text{isobaric}}={\left(\Delta Q\right)}_{\text{isochoric}}$ ⇒ $\mu C_p{\left(\Delta T\right)}_A=\mu C_v{\left(\Delta T\right)}_B$

⇒ ${\left(\Delta T\right)}_B=\frac{C_p}{C_v}{\left(\Delta T\right)}_A=\gamma {\left(\Delta T\right)}_A=1.4\times 30=42K.$

According to the NCERT text, 'A reversible process proceeds infinitely slowly by a series of equilibrium states such that system and the surroundings are always in near equilibrium with each other.' It also states that 'A process is reversible only if it is quasi-static (...) and there are no dissipative effects.' Therefore, the presence of significant dissipative effects is not characteristic of a reversible process.

The NCERT text states, 'From the preceding discussion, a reversible process is an idealised notion. A process is reversible only if it is quasi-static (system in equilibrium with the surroundings at every stage) and there are no dissipative effects.' These ideal conditions are rarely met in real-world processes, making it an idealised concept.

The NCERT text explicitly states, 'The base of a vessel on an oven is hotter than its other parts. When the vessel is removed, heat is transferred from the base to the other parts, bringing the vessel to a uniform temperature... The process cannot be reversed; a part of the vessel will not get cooler spontaneously and warm up the base.' Options o1, o3, and o4 describe characteristics or examples of reversible processes.