NEET Practice Questions (MCQs) with Answers & Solutions

For an ideal gas, the work done in an isothermal compression (W₁) is less than the work done in an adiabatic compression (W₂) for the same volume change. This is because in the adiabatic case, the gas also gains internal energy due to the work done against intermolecular forces, making W₂ greater than W₁.

$$\begin{gathered} V{p}^2=cons\tan t \\ V\frac{n^2{R}^2{T}^2}{V^2}=Cons\tan t (U\sin g Ideal Gas Eqn) \\ {T}^2 \alpha V \end{gathered}$$

(b) For a refrigerator, we know that 

             $\frac{Q_1}{W}=\frac{Q_1}{Q_1-Q_2}=\frac{T_1}{T_1-T_2}$

where,

$Q_1$=amount of heat delivered to the room

W = electrical energy consumed

$T_1$= room temperature= $t_1+273$

$T_2=$temperature of sink=$t_2+273$

$\therefore$   $\frac{Q_1}{1}=\frac{t_1+273}{t_1+273-\left(t_2+273\right)}$

$\Rightarrow Q_1=\frac{t_1+273}{t_1-t_2}$

(b) Given temperature of source T=30°C=30+273 T1=303K
Temperature of sink T₂=4°C=4+273 T₂=277K

As we know that 
Q₁/Q₂=T₁/T₂
=>Q₂+W/Q₂=T₁/T₂ ..........(W=Q₁-Q₂)

where Q₂ is the amount of heat drawn from the sink (at T₂),W is workdone on working substance,
Q₁ is amount of heat rejected to source (at room temperature $T_1$).

=>WT₂+T₂Q₂=T₁Q₂
=>WT₂=T₁Q₂-T₂Q₂
=>WT₂=Q₂(T₁-T₂)
=>W=Q₂(T₁/T₂-1)
=>W=600X4.2X(303/277-1)
W=600X4.2X(26/277)
W=236.5Joules

Power=Workdone/Time=W/t=236.5/1=236.5W

Key Concept:

Coefficient of performance (β) of a refrigerator is defined as the ratio of quantity of heat removed per cycle  to the work done on the working substance per cycle to remove this heat.

Given, coefficient of performance of a refrigerator β=5

Temperature of surface i.e. inside freezer,

T₂=-20°C=-20+273=253K

Temperature of surrounding i.e. heat rejected outsider T₁=?

So,β=T₂/T₁-T₂

$5=\frac{253}{T_1-253}$

<sub>$5T_1$=$1518T_1$=$\frac{1518}{5}$=$303.6K$

$T_1=303.6-273=31°C$</sub>

$$\begin{gathered} \left(\mathrm{d}\right) \\ \mathrm{As} \mathrm{per} \mathrm{the} \mathrm{question}; \\ \mathrm{P}\propto {\mathrm{T}}^3 \\ \mathrm{As} \mathrm{we} \mathrm{know}, \mathrm{PV}=\mathrm{nRT} \\ \Rightarrow \mathrm{P}\propto {\left(\mathrm{PV}\right)}^3 \\ \Rightarrow {\mathrm{P}}^2{\mathrm{V}}^3=\mathrm{constant} \\ \Rightarrow {\mathrm{PV}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\mathrm{Constant} \\ \Rightarrow \mathrm{\gamma }=\frac{3}{2}\Rightarrow \frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}=\frac{3}{2} \end{gathered}$$

$$\begin{gathered} {\mathrm{T}}_1=273+27=300 \mathrm{k} \\ {\mathrm{T}}_2=273+927=1200 \mathrm{k} \end{gathered}$$ 

Gas equation for adiabatic process

         ${\mathrm{PV}}^{\mathrm{\gamma }}=$constant 

$$\begin{gathered} \mathrm{P}{\left(\frac{\mathrm{T}}{\mathrm{P}}\right)}^{\mathrm{\gamma }}= \mathrm{constant} (\because \mathrm{PV}=\mathrm{nRT}) \\ {\mathrm{P}}^{1-\mathrm{\gamma }}=\frac{\mathrm{C}}{{\mathrm{T}}^{\mathrm{\gamma }}}\Rightarrow \frac{{{\mathrm{P}}_1}^{{}^{1-\mathrm{\gamma }}}}{{{\mathrm{P}}_2}^{{}^{1-\mathrm{\gamma }}}}=\frac{{{\mathrm{T}}_2}^{\mathrm{\gamma }}}{{{\mathrm{T}}_1}^{\mathrm{\gamma }}}\Rightarrow {\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)}^{\mathrm{\gamma }-1}={\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^{\mathrm{\gamma }} \\ \therefore \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}={\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^{\frac{\mathrm{\gamma }}{\mathrm{\gamma }-1}} \\ 0\mathrm{r} {\mathrm{P}}_2={\mathrm{P}}_1{\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^{\frac{\mathrm{\gamma }}{\mathrm{\gamma }-1}} \\ {\mathrm{P}}_2=2{\left(\frac{1200}{300}\right)}^{\frac{1.4}{1.4-1}} \\ {\mathrm{P}}_2=256 \mathrm{atm} \end{gathered}$$

$$\begin{gathered} {\mathrm{p}}_1{\mathrm{V}}_1^{5/3}=\mathrm{p}\text{'}{\left(\frac{{\mathrm{V}}_1}{8}\right)}^{5/3} \\ \mathrm{p}\text{'}={\mathrm{p}}_1(8{)}^{5/3} \\ ={\mathrm{p}}_1\times 2^5 \\ \mathrm{p}\text{'}=32 {\mathrm{p}}_1 \end{gathered}$$