Thermodynamics Practice Questions

${\text{W}}_{ {\text{DE}}}=\;\frac{1}{2}(600+300) 3 {\text{J}}$ $=1350 {\text{J}}$ ${\text{W}}_{ {\text{EF}}}=-300 \times 3=-900 {\text{J}}$ ${\text{W}}_{ {\text{DEF}}}=450 {\text{J}}$

${[ {\text{P}}_{\;\text{avg }\;}=0]}$ (due to random motion) ${\text{v}}_{\;\text{rms }\;}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}}}$ ${\text{T}}_{\;\text{new }\;}=2 {\text{T}}$ ${\text{M}}_{\;\text{new }\;}=\;{ {\text{M}}}/{2}$ $\;{ {\text{v}}_{\;\text{new }\;}}/{ {\text{v}}}=\;{\sqrt{{2 {\text{T}}}/{ {\text{M}} / 2}}}/{\sqrt{\;{ {\text{T}}}/{ {\text{M}}}}}$ ${\text{v}}_{\;\text{new }\;}=2 {\text{v}}$

$Q=m s \Delta T$ $\;\frac{d Q}{d t}=(\;\frac{d m}{d t})_{\;\text{water }\;} S \Delta T=(\;\frac{d m}{d t})_{\;\text{oil }\;} C$ $\Rightarrow 2 \times 4.2 \times 10^3 \times 40=(\;\frac{d m}{d t})_{o i l} \times 8 \times 10^6$ $\Rightarrow (\;\frac{d m}{d t})_{o i l}=\;\frac{8 \times 4.2 \times 10^4}{8 \times 10^6} {\text{kg}} / {\text{minute}}$ $=42 {\text{g}} / {\min}$

$v_{r m s}=\sqrt{\;\frac{3 R T}{M}}$ $v_p=\sqrt{\;\frac{2 R T}{M}}$ $\Rightarrow v_{r m s}=\sqrt{\;\frac{3}{2}} v_p$

$\Delta L_{1}=\alpha _{1} L_{1} \Delta T$ $\Delta L_{2}=\alpha _{2} L_{2} \Delta T$ $\alpha _{1} L_{1}=\alpha _{2} L_{2}$ $1.2 \times 10^{-5} \times L_{1}=1.8 \times 10^{-5} \times L_{2}$ $L_{1}=\;\frac{1.8}{1.2} \times 40=60 {cm}$

From the equation{[a] \equiv \[P V^2]$ }$${[b] \equiv [V]}$$\Rightarrow [;\frac{a}{b}] \equiv [P V]$

Total number of moles are$n=n_{H_2}+n_{O_2}$ $=\;\frac{16}{2}+\;\frac{128}{32}$ $=12 \;\text{ moles }\;$ $\;\text{ Using }\; P V=n R T$ $V=\;\frac{n R T}{P}$ $=\;\frac{12 \times 8.31 \times 273.15}{10^5} {\text{m}}^3$ $=0.27 {\text{m}}^3=27 \times 10^4 {\text{cm}}^3$

${\text{W}}= {\text{nR}} \Delta {\text{T}}=400 {\text{J}}$ $\therefore \Delta {\text{Q}}= {\text{nC}}_{ {\text{P}}} \Delta {\text{T}}$ $=n \times \;\frac{7}{2} R \times \Delta T=\;\frac{7}{2} \times (400)=1400$

Let $A B$ is isothermal process and $B C$ is adiabatic process then for $A B$ process${P}_{ {A}} {V}_{ {A}}= {P}_{ {B}} {V}_{ {B}}$ $\Rightarrow {P}_{ {B}}=10^{7} {Nm}^{-2}$For process BC${P}_{ {B}} {V}_{B}^{r}= {P}_{ {C}} {V}_{C}^{r}$ $P_{C}=3.536 \times \times 10^{6} {Pa}$

$C_V=\;\frac{f}{2} R$total degree of freedoms$=1 \times 3+3 \times 5=18$ $\;\frac{\alpha ^2}{4}=\;\frac{18}{2 n}=\;\frac{18}{2 \times 4}$ $\Rightarrow \alpha ^2=9$ $\alpha =3$

Initially : $\;\frac{1}{4}=1-\;\frac{300}{T_H}$ $\Rightarrow T_H=400 K$Finally : Efficiency becomes $\;\frac{1}{2}$ $\Rightarrow \;\frac{1}{2}=1-\;\frac{300}{T_H^{'}}$ $\Rightarrow T_H^{'}=600 K$ $\Rightarrow$ Temperature of the source increases by $200^{\circ} {\text{C}}$.

$η=1-\;{ {\text{T}}_{ {\text{L}}}}/{ {\text{T}}_{ {\text{H}}}}$ $\;\frac{1}{2}=1-\;{ {\text{T}}_{ {\text{L}}}}/{ {\text{T}}_{ {\text{H}}}}$ $\;\frac{1}{2}(1 \cdot 3)=1-(\;{ {\text{T}}_{ {\text{L}}}-40}/{ {\text{T}}_{ {\text{H}}}})$ $\;\frac{1}{2}(1 \cdot 3)=\;\frac{1}{2}+\;{40}/{ {\text{T}}_{ {\text{H}}}} \;\; {\text{T}}_{ {\text{H}}}=266.7 {\text{K}}$

Statement A is incorrect, statement B is correct by equipartition of energy. Statement C is incorrect as monoatomic does not have any rotational degree of freedom and CH4 is a polyatomic gas so it has 6 degree of freedom. So only B and D are correct.

According to kinetic theory of gases,A. The motion of the gas molecules freezes at $0 {\text{K}}$.B. The mean free path decreases on increasing the number density of the molecules as$\mu =\;{1}/{{\sqrt{2}} \pi n d^2} \Rightarrow \mu \propto \;\frac{1}{n} \;\text{. }\;$C. The mean free path increases on increasing the volume. Now if temperature is increased by keeping the pressure constant the volume should increase that is mean free path increases.D. K.E.avg per molecule per degree of freedom is $\;\frac{1}{2} k_B T$.

No of moles $=\;\frac{44.8}{22.4}=2$Gas is mono atomic so ${\text{C}}_{ {\text{v}}}=\;\frac{3}{2} {\text{R}}$ $\Delta {\text{Q}}= {\text{nC}}_{ {\text{v}}} \Delta {\text{T}}$ $=2 \times \;\frac{3}{2} {\text{R}}(20)$ $=60 {\text{R}}$ $=60 \times 8.3$ $=498 {\text{J}}$

Molar mass $M-\;\frac{2 \times 4+n \times 1}{2+n} . . . . .$ (i)Also, $\gamma =\;{n_{1} C_{P_{1}}+n_{2} C_{P_{2}}}/{n_{1} C_{V_{1}}+n_{2} C_{V_{2}}}=\;\frac{2 \times 5 R+n \times 7 R}{2 \times 3 R+n \times 5 R}$ $\Rightarrow \gamma =\;\frac{10+7 n}{6+5 n} . . . . . . \;\text{ (ii) }\;$Given that $V_{r m s}=\sqrt{2} V_{\;\text{sound }\;}$ $\Rightarrow \sqrt{\;\frac{3 R T}{M}}=\sqrt{2} \sqrt{\;\frac{\gamma R T}{M}}$ $\Rightarrow \gamma =\;\frac{3}{2}$ $\Rightarrow n=2$

$η=\;\frac{W}{Q}=\;\frac{300-225}{300}$ $\Rightarrow \;\frac{75}{300}=1-\;\frac{T_L}{T_H}$ $\Rightarrow T_L=\;\frac{3}{4} T_H=\;\frac{3}{4}(500)=375 {\text{K}}$ $\Rightarrow T_L=102^{\circ} {\text{C}}$

Here volume is constant.$\therefore \;\frac{P_1}{T_1}=\;\frac{P_2}{T_2}$ $\Rightarrow \;\frac{100}{273}=\;\frac{180}{T_2}$ $\Rightarrow T_2=\;\frac{180}{100} \times 273$ $=1.8 \times 273$ $=491 {\text{K}}$