Thermodynamics MCQs for NEET — Physics Questions with Answers

Volume of the gas $V=\frac{m}{d}$ and using $P{V}^{\gamma }$ = constant

We get $\frac{P\text{'}}{P}={\left(\frac{V}{V\text{'}}\right)}^{\gamma }=\hspace{0.17em}{\left(\frac{d\text{'}}{d}\right)}^{\gamma }={\left(32\right)}^{7/5}=128$

$\frac{T_2}{T_1}={\left(\frac{{\displaystyle V_1}}{{\displaystyle V_2}}\right)}^{\gamma -1}\Rightarrow T_2=300\hspace{0.17em}{\left(\frac{{\displaystyle 27}}{{\displaystyle 8}}\right)}^{\frac{5}{3}-1}=300\hspace{0.17em}{\left(\frac{{\displaystyle 27}}{{\displaystyle 8}}\right)}^{\frac{2}{3}}$

$\text{=300\hspace{0.17em}}{\left\{{\left(\frac{{\displaystyle 27}}{{\displaystyle 8}}\right)}^{1/3}\right\}}^2=300\hspace{0.17em}{\left(\frac{{\displaystyle 3}}{{\displaystyle 2}}\right)}^2=675K$

$\Rightarrow \hspace{0.17em}\Delta T=675-300=375\hspace{0.17em}K$

For the expansion of an ideal gas, the work done is greater in the isothermal process compared to the adiabatic process. In the isothermal process, heat is continuously absorbed from the surroundings, allowing for more expansion and thus more work done.

Since PV = RT and T = constant;

∴ PV = constant.

For Isothermal process PV = constant

$\Rightarrow \left(\frac{{\displaystyle dP}}{{\displaystyle dV}}\right)=\frac{-P}{V}=$ Slope of Isothermal curve

For adiabatic $P{V}^{\gamma }=$ constant

$\Rightarrow \frac{dP}{dV}=\frac{-\gamma P}{V}=$ Slop of adiabatic curve slope

Clearly, ${\left(\frac{{\displaystyle dP}}{{\displaystyle dV}}\right)}_{\text{adiabatic}}=\gamma {\left(\frac{{\displaystyle dP}}{{\displaystyle dV}}\right)}_{\text{Isothermal }}$

$dQ=0=-2+dW\Rightarrow dW=2J$

⇒ Work done by the gas = 2 J

Slope of adiabatic curve = $\gamma$ × (Slope of isothermal curve)

$$\begin{gathered} \mathrm{Isothermal} \mathrm{proccess} : \mathrm{PV}=\mathrm{constant} \\ \mathrm{Differentiating}, \mathrm{we} \mathrm{get} \mathrm{PdV}+\mathrm{VdP}=0 \\ \Rightarrow \mathrm{Slope} \mathrm{of} \mathrm{isothermal} \mathrm{curve} {\left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)}_{\mathrm{iso}} =\frac{-\mathrm{P}}{\mathrm{V}} .........\left(1\right) \\ \mathrm{Adiabtic} \mathrm{process}: {\mathrm{PV}}^{\mathrm{\gamma }}=\mathrm{constant} \\ \mathrm{Differentiating}, \mathrm{we} \mathrm{get} {\mathrm{P\gamma V}}^{\mathrm{\gamma }-1} \mathrm{dV}+{\mathrm{V}}^{\mathrm{\gamma }} \mathrm{dP}=0 \\ \Rightarrow \mathrm{Slope} \mathrm{of} \mathrm{adiabatic} \mathrm{curve} {\left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)}_{\mathrm{adi}} = \frac{-\mathrm{\gamma P}}{\mathrm{V}} ............\left(2\right) \\ \Rightarrow \mathrm{Ratio} \mathrm{of} \mathrm{slopes} \frac{{\left({\displaystyle \frac{\mathrm{dP}}{\mathrm{dV}}}\right)}_{\mathrm{adi}}}{{\left({\displaystyle \frac{\mathrm{dP}}{\mathrm{dV}}}\right)}_{\mathrm{iso}}}=\frac{\left(-\mathrm{\gamma P}/\mathrm{V}\right)}{-\mathrm{P}/\mathrm{V}}=\mathrm{\gamma } \end{gathered}$$

Due to sudden  compression  (adiabatic) ,the temperature of the system increases to a very high value. This causes the flow of heat from system to the surroundings, thus decreasing the temperature. This decrease in temperature results in decrease in pressure.

Adiabatic Bulk modulus $E_{\varphi }=\gamma P$

In adiabatic process, no heat transfers between system and surrounding.