Thermodynamics Practice Questions

$\Delta D=D \alpha \Delta T$ $\Delta T=\;\frac{0.011}{6.230 \times 1.4 \times 10^{-5}}$ $=126.11^{\circ} {\text{C}}$ $\Rightarrow T_f= {\text{T}}+\Delta T$ $=(27+126.11)^{\circ} {\text{C}}$ $=153.11^{\circ} {\text{C}}$

$U=2 (\;\frac{3}{2} R) 300$ $=3 \times 8.31 \times 300$ $=7479 {\text{J}}$

First case $: η=1-\;\frac{100}{300}=\;\frac{2}{3}$Second case $: η_{\;\text{net }\;}=η_1+η_2-η_1 η_2$ $η_1=1-\;\frac{200}{300}=\;\frac{1}{3}$ $η_2=1-\;\frac{100}{200}=\;\frac{1}{2}$ $η_{\;\text{net }\;}=\;\frac{1}{3}+\;\frac{1}{2}-\;\frac{1}{6}=\;\frac{2}{3}$ $η($ first case $)=η($ second case $)$

Because the vessel is closed, it will be an isochoric process.To double the speed, temperature must be 4 times $(∨ \alpha \sqrt{T})$So, $T_f=1600 {\text{K}}, T_i=400 {\text{K}}$number of moles are $\;\frac{56}{28}=2$ $\;\text{ so }\; Q=n C v \Delta T=2 \times \;\frac{5}{2} \times 2 \times 1200$ $=12000 {\text{cal}}=12 {\text{K}} {\text{cal}}$

$\;\frac{C_P}{C_V}=\gamma$ $C_V=(\;\frac{f}{2}) R$ and $C_P-C_V=R$ $\Rightarrow \;\frac{C_P}{C}=\;\frac{1+f / 2}{f / 2}=1+\;\frac{2}{f}$

$H=\;\frac{V^2}{R} \cdot \Delta t$ $\Rightarrow H=\;\frac{V^2}{R_1} \cdot 20=\;\frac{V^2}{R_2} \cdot 60 . . . .$. (i)Also, $H=\;\frac{V^2}{ [\;\frac{R_1 R_2}{R_1+R_2}] } \cdot \Delta t$ $=\;\frac{4}{3} \cdot \;\frac{V^2}{R_1} \cdot \Delta t [ \because R_2=3 R_1]$ $\Rightarrow \Delta t=15$

As per ideal gas equation, $V=\;\frac{n R}{P} T$ $\Rightarrow$ Slope of V-T graph is inversely proportional to ${\text{P}}$.As $m_2>m_1 \Rightarrow P_1>P_2$

$P V=n R T$So $\;\frac{d P}{P} \times 100=\;\frac{d T}{T} \times 100$ $0.4=\;\frac{1}{T} \times 100$ $\Rightarrow T=250 {\text{K}}$

$(1-\;\frac{324}{T_H})=\;\frac{300-180}{300}$ $1-\;\frac{2}{5}=\;\frac{324}{T_H}$ $T_H=\;\frac{324 \times 5}{3}=540$

$η=1-\;\frac{T_2}{T_1}$ $=1-\;\frac{200}{800}=\;\frac{3}{4}$ $\therefore η=\;\frac{W}{Q_1}$ $\Rightarrow \;\frac{3}{4}=\;\frac{12000 \times 10^3}{Q_1}$ $\Rightarrow Q_1=16 \times 10^6 {\text{J}}$

$K E_{a v g}=\;\frac{3}{2} k T$ (At lower temperature)As temperature is same for both the gases.$\Rightarrow$ Both gases will have same average kinetic energy.$\Rightarrow \;{(K E_{\;\text{avg }\;})_{\;\text{arg on }\;}}/{(K E_{\;\text{avg }\;})_{\;\text{oxygen }\;}}=\;\frac{1}{1}$

$η=1-\;\frac{T_2}{T_1}=1-\;\frac{273}{373}$ $=0.26809$ $\approx 26.81 \%$

A. ${\text{V}}_{ {\text{Rms}}}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}_{ {\text{w}}}}} \Rightarrow {\text{V}}_{ {\text{Rms}}}$ is sameB. $\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;{ {\text{N}}_1}/{ {\text{N}}_2} \Rightarrow {\text{B}}$ is correct

$P_1 V_1^\gamma =P_2 V_2^2$ $\;\frac{P_1}{d_1^\gamma }=\;\frac{P_2}{d_2^\gamma }$ $\;\frac{d_1 T_1}{d_1^\gamma }=\;\frac{d_2 T_2}{d_2^\gamma }$ $T_2= (\;\frac{d_2}{d_1}) ^{\gamma -1} T_1$ $=(32)^{\;\frac{2}{5}} T_1$ $T_2=4 T_1$

$W=\;\frac{\mu R (T_2-T_1) }{1-r}$ for a polytropic process for adiabatic process ${\text{r}}=\gamma$ $\Rightarrow$ Statement I is true.In an adiabatic process$\Delta {\text{U}}=-\Delta {\text{W}}$ $\Rightarrow$ If work is done on the gas$\Rightarrow \Delta {\text{W}}$ is negative$\Rightarrow \Delta U$ is positive or temperature increases$\Rightarrow$ Statement II is true

Given, atomic mass of monoatomic gas, $m=4 u$ Velocity of container, $u=30 {ms}^{-1}$ Final velocity of container, $v=0 {ms}^{-1}$ Let $\Delta T$ be the change in temperature and $\Delta U$ be the internal energy change. Therefore, $\Delta {KE}=\Delta U$ $\;\; =(1 / 2) m(v^{2}-u^{2})=n C_{v} \Delta T$ $\Delta T=\;\frac{1}{2} \;\frac{m(v^{2}-u^{2})}{n \cdot 3 / 2 R}$ $\Rightarrow \;\; \;\; \;\; \;\; \;\; \;\; ( \because C_{v}=\;\frac{3}{2} R)$ \;\; \;\frac{1}{2} \times m\[v^{2}-u^{2}]$=n \cdot ;\frac{3}{2} \cdot R \Delta T$$;\text{ Now, }; ;\frac{1}{2} \times 4|0^{2}-30^{2}|=1 \times ;\frac{3}{2} R \Delta T$$\Delta T=;\frac{4 \times 900}{3 R}=;\frac{x}{3 R}$$;; x=3600$

Given, the average translational kinetic energy of dinitrogen(\text{N}_2$) = Kinetic energy of an electron ...(i) Translational kinetic energy of dinitrogen (\text{N}_2$ ) ${\text{KE}}=\frac{3}{2} \text{K}_{\text{B}} \text{T}$ Here, T = temperature of the gas, and $\text{K}_\text{B}$ = Boltzmann constant. Kinetic energy of an electron = eV Given, the potential difference of an electron, V = 0.1 V Substituting the values in the Eq. (i), we get $\frac{3}{2} \text{K}_{\text{B}}\ \text{T}={\text{eV}}$ \Rightarrow $\frac{3}{2} \times 1.38 \times 10^{-23} \times \text{T}=1.6 \times 10^{-19} \times (0.1)$ $\text{T}=773 {\text{K}}=773-273° {\text{C}}=500° {\text{C}}$

According to first law of thermodynamics dQ = dU + dW.......(i) As we know that, work done by the gas depends on the type of process, i.e. path and $d U$ depends on the initial and final states. So, considering Eq. (i), dQ will also be dependent on path. It means that, in thermodynamics, heat and work are path functions.