Thermodynamics Practice Questions

$A$ to $B$ is an isobaric process. The work done$W=n R (T_2-T_1) =2 R(500-300)=400 R$

Volume $=\;\frac{\;\text{ mass }\;}{\;\text{ density }\;}=\;\frac{1}{4} m^3$ $K . E=\;\frac{5}{2} P V$ $=\;\frac{5}{2} \times 8 \times 10^4 \times \;\frac{1}{4}$ $=5 \times 10^4 J$

Work done by the system in the isothermal process$D A$ is $W=2.303 n R T \log _{10} \;\frac{P_D}{P_A}$ $=2.303 \times 2 R \times 300 \log _{10} \;\frac{1 \times 10^5}{2 \times 10^5}=-414 R$.Therefore work done on the gas is $+414 R$.

Same as $A .20$

The speed of sound in a gas is given by $v=\sqrt{\;\frac{\gamma R T}{M}}$ $\; \therefore \;\frac{v_{O_2}}{v_{H e}}=\sqrt{\;\frac{\gamma _{O_2}}{M_{O_2}}} \times \;\frac{M_{H e}}{\gamma _{H e}}$ $\;=\sqrt{\;\frac{1.4}{32} \times \;\frac{4}{1.67}}=0.3237$ $\; \therefore v_{H e}=\;\frac{v_{O_2}}{0.3237}$ $\;=\;\frac{460}{0.3237}$ $\;=1421 {\text{m}} / {\ s}$

$\;\frac{K_1 A (T_1-T) }{ℓ_1}=\;\frac{K_2 A (T-T_2) }{ℓ_2}$ $\therefore T=\;\frac{K_1 T_1 ℓ_2+K_2 T_2 ℓ_1}{K_2 ℓ_1+K_1 ℓ_2}$ $=\;\frac{K_1 ℓ_2 T_1+K_2 ℓ_1 T_2}{K_1 ℓ_2+K_2 ℓ_1}$

The efficiency $(η)$ of a Carnot engine and the coefficient of performance $(\beta )$ of a refrigerator are related as $\beta =\;\frac{1-η}{η}$ $\;\;$ Here, $η=\;\frac{1}{10}$ $\;\; \therefore \beta =\;\frac{1-\frac{1}{10}}{ (\;\frac{1}{10}) }=9$ Also, Coefficient of performance $(\beta )$ is given by $\beta =\;\frac{Q_2}{W}$, where $Q_2$ is the energy absorbed from the reservoir.or, $9=\;\frac{Q_2}{10}$ $\therefore Q_2=90 {\text{J}} \;\text{. }\;$

According to Mayer's relationship $C_p-C_v=R$ $\therefore \;\frac{C_p}{M}-\;\frac{C_v}{M}=\;\frac{R}{M} \;\; \;\text{ Here }\; M=28$

For path iaf, $\Delta U=Q-W=50-20=30 {\text{cal}}$. For path ibf, $W=Q-\Delta U=36-30=6 {\text{cal}}$.

Heat lost by ${\text{He}}=$ Heat gained by $N_2$ $\;n_1 C v_1 \Delta T_1=n_2 C v_2 \Delta T_2$ $\;\;\frac{3}{2} R [\;\frac{7}{3} T_0-T_f]$ $\;=\;\frac{5}{2} R [T_f-T_0] \Rightarrow T_f$ $\;=\;\frac{3}{2} T_0$

$\;W=\;\frac{n R \Delta T}{1-\gamma } \Rightarrow -146000$ $\;=\;\frac{1000 \times 8.3 \times 7}{1-\gamma }$ $\;\;\text{ or }\; 1-\gamma =-\;\frac{58.1}{146} \Rightarrow \gamma$ $\;=1+\;\frac{58.1}{146}=1.4$Hence the gas is diatomic.

Total power radiated by Sun $=\sigma T^4 \times 4 \pi R^2$The intensity of power at earth's surface $=\;\frac{\sigma T^4 \times 4 \pi R^2}{4 \pi r^2}$Total power received by Earth $=\;\frac{\sigma T^4 R^2}{r^2} (\pi r_0^2)$

Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states $i . e ., \Delta U_1=\Delta U_2$

$\;\frac{n_1+n_2}{r-1}=\;\frac{n_1}{r_1-1}+\;\frac{n_2}{r_2-1}$ $\;\frac{\;\frac{r}{4}+\frac{16}{32}}{r-1}=\;\frac{16 / 4}{\frac{5}{3}-1}+\;\frac{16 / 32}{1.4-1}$ $\therefore \gamma =1.62$

Consider a shell of thickness $(d r)$ and of radius $(r)$ and the temperature of inner and outer surfaces of this shell be $T,(T-d T)$ $\;H=\;\frac{K A[(T-d T)-T]}{d r}=\;\frac{-K A d T}{d r}$ $\;H=-4 \pi K r^2 \;\frac{d T}{d r}$ as $A=4 \pi r^2)$Then, $(H) ∫_{r_1}^{r^2} \;\frac{1}{r^2} d r=-4 \pi K ∫_{T_1}^{T_2} d T$ $H [\;\frac{1}{r_1}-\;\frac{1}{r_2}] =-4 \pi K [T_2-T_1]$or $H=\;\frac{-4 \pi \vec{K} r_1 r_2 (T_2-T_1) }{ (r_2-r_1) }$

$(b, c)$ are wrong. As first law is applicable to a cyclic process and concept of entropy is introduced by the second law.

$Q_1=T_0 S_0+\;\frac{1}{2} T_0 S_0=\;\frac{3}{2} T_0 S_0$ $Q_2=T_0 (2 S_0-S_0) =T_0 S_0$and $Q_3=0$ $η=1-\;\frac{Q_2}{Q_1}=1-\;\frac{T_0 S_0}{\frac{3}{2} T_0 S_0}=\;\frac{1}{3}$

The thermal resistance$\;\;\frac{x}{K A}+\;\frac{4 x}{2 K A}=\;\frac{3 x}{K A}$ $\; \therefore \;\frac{d Q}{d t}=\;\frac{\Delta T}{\frac{3 x}{K A}}=\;\frac{ (T_2-T_1) K A}{3 x}$ $\;=\;\frac{1}{3} \{\;\frac{A (T_2-T_1) K}{x}\}$ $\; \therefore f=\;\frac{1}{3}$

Here $Q=0$ and $W=0$. Therefore from first law of thermodynamics $\Delta U=Q+W=0$ $\therefore$ Internal energy of the system with partition $=$ Internal energy of the system without partition.$\;n_1 C_v T_1+n_2 C_v T_2= (n_1+n_2) C_v T$ $\; \therefore T=\;\frac{n_1 T_1+n_2 T_2}{n_1+n_2}$But $n_1=\;\frac{P_1 V_1}{R T_1}$ and $n_2=\;\frac{P_2 V_2}{R T_2}$ $\; \therefore T=\;\frac{\;\frac{P_1 V_1}{R T_1} \times T_1+\frac{P_2 V_2}{R R_2} \times T_2}{\frac{P_1 V_1}{R R_1}+\;\frac{R_2 V_2}{R R R_2}}$ $\;=\;\frac{T_1 T_2 (P_1 V_1+P_2 V_2) }{P_1 V_1 T_2+P_2 V_2 T_1}$

$\Delta Q=m C \times \Delta T$ $=1 \times 4180 \times (40-10)=80 \times 30$( $\therefore \Delta Q=$ heat supplied in time $t$ for heating $1 L$ water from $10^{\circ} {\text{C}}$ to $40^{\circ} {\text{C}}$ ) also $\Delta Q=836 \times t \Rightarrow t=\;\frac{4180 \times 30}{836}=150 {\ s}$