Q: Derive an expression for work done in adiabatic expansion

Adiabatic process : \$PV^{\gamma} = K\$So, \$P = KV^{-\gamma}\$Work done \$W = \int PdV\$Or \$W = \int KV^{-\gamma} dV\$Or \$W = K \times \dfrac{V^{-\gamma+1}}{1-\gamma}\bigg|_{V_1}^{V_2}\$Or \$W = \dfrac{K}{1-\gamma}\times [{V_2^{-\gamma+1}-V_1^{-\gamma+1}}]\$Or \$W = \dfrac{1}{1-\gamma}\times [{KV_2^{-\gamma+1}-KV_1^{-\gamma+1}}]\$Or \$W = \dfrac{1}{1-\gamma}\times [P_2V_2^{\gamma}{V_2^{-\gamma+1}-P_1V_1^{\gamma}V_1^{-\gamma+1}}]\$Or \$W = \dfrac{P_2V_2-P_1V_1}{1-\gamma}\$

Q: During an adiabatic process, the pressure of a gas is found to the proportional to the cube of its temperature. The ratio of \$\displaystyle \frac{C_{p}}{C_{v}}\$ for the gas is

\$P\propto T^{3}\$,\$PV=nRT\$\$P\propto T^{3}\$\$P\propto (PV)^{3}\$\$P^{2}V^{3}=constant\$\$\displaystyle PV^{\dfrac{3}{2}}=constant\$\$\displaystyle \gamma =\dfrac{3}{2}\$

Q: A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step of the staircase is 15 cm, find the power of the boy.

Given mass of body \$(m)=50kg\$\$h=45\times 15=675cm=6.75m\$\$t=9s\$ \$\delta =10m\$Potential energy \$=mgh\$ \$=\left( 50\times 10\times 6.75 \right) 5\$ \$=33755\$Power of the body \$=\dfrac { Potential\quad energy }{ Time } \$ \$=\dfrac { 3375 }{ 9 } \$ \$=375W\$

Q: A carnot engine working between 300 K and 600 K has work output of 800 J per cycle. The amount of heat energy supplied to the engine from source per cycle will be:

Given : For Carnot engine,\$\Rightarrow T_1=600K\$ \$T_2=300K\$\$\Rightarrow W=800J\$We know,Efficiency \$(n)=1-\dfrac{T_2}{T_1}=\dfrac{W}{Q}\$\$\Rightarrow \dfrac{W}{Q}=1-\dfrac{300}{600}\$\$\Rightarrow Q=\dfrac{W}{1-\dfrac{1}{2}}\$\$\Rightarrow \dfrac{800}{\dfrac{1}{2}}\$\$\Rightarrow Q=1600J\$i.e, heat given to system is \$1600J.\$Hence, the answer is \$1600.\$

Q: An ideal gas heat engine operates in carnot cycle between \${227}^{o}C\$ and \${127}^{o}C\$. It absorbs \$6 \times {10}^{4} cal\$ of heat at higher temperature. Amount of heat converted into work, is

For a Carnot engine, efficiency \$e = 1 - \frac{T_2}{T_1}= 1- \frac{127 + 273}{227 + 273} = 1 - \frac{400}{500} = \frac{1}{5}\$Now, \$e = \frac{Work \space output}{Heat \space input} = \frac{W}{6 \times 10^{4}}\$\$W = e \times 6 \times 10^{4} = \frac{1}{5} \times 6 \times 10^{4} = 1.2 \times 10^{4} cal\$

Q: In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas if the initial internal energy of the gas was 30 J. The final internal energy wilI be:

dQ = -20J, dW = -8 J, \$U_1 = 30J\$By 1st law of thermodynamics,dQ = dU + dW\$\Rightarrow \$ dU = dQ-dW= -20 + 8 = -12JFinal internal energy, \$U_2 = U_1 + dU\$= 30-12 = 18 J

Q: In a given process on an ideal gas \$dW = 0\$ and \$dQ < 0\$, then for the gas :

From first law of thermodynamics,\$dQ=dU+dW\$we have \$dQ=dU\$ \$\left( as\quad dW=0 \right) \$But \$dQ < 0\$\$\therefore dU < 0\$\$N{ C }_{ V }\Delta T< 0\$or \$\Delta T < 0\$Hence, the temperature will decrease.

Q: Two Carnot engines A and B are operated in series. The first one A receives heat at 800 K and rejects to a reservoir at temperature T. The second engine B receives the heat rejected by the first engine and in turn rejects to a heat reservoir at 300 K. Calculate the temperature T for the following situations:(a) The outputs of the two engines are equal and(b) The efficiencies of two engines are equal

Given:\$T_1=800K\$\$T_3=400K\$To find:Temperature T (i) when output of two engines are equal(ii) when efficiencies of two engines are equal.Let the output of both engines be W.Let the engine A take \$Q_1\$ heat as input at temperature \$T_1\$ and gives out heat \$Q_2\$ at temperature \$T\$ the second engine B receive \$Q_2\$ as input and give out \$Q_3\$ at temperature \$T_2\$ to the sink.Work done by engine, A \$W = Q_1 – Q_2\$Work done by engine, B \$W = Q_2 – Q_3\$Thus, \$Q_1 – Q_2 = Q_2 – Q_3\$Dividing both sides by \$Q_1\$,\$1 – \dfrac {Q_2}{Q_1} = \dfrac {Q_2}{Q_1} – \dfrac {Q_3}{Q_1}\$\$\implies 1-\dfrac T{T_1} = \dfrac {Q_2}{ Q_1}\left (1- \dfrac {Q_3}{ Q_2}\right)\$\$\implies 1-\dfrac T{T_1} = \dfrac {Q_2}{ Q_1}\left (1- \dfrac {T_3}{ T}\right)\$\$\implies 1-\dfrac T{T_1} = \dfrac T{T_1} \left(1- \dfrac {T_3} T\right)\$\$\implies \dfrac {T_1}T – 1 = 1 - \dfrac {T_3} T\$\$\implies \dfrac {T_1}T + \dfrac {T_3} T = 2\$\$\implies \dfrac 1T (T1 + T3) = 2\$\$\implies T =\dfrac {(T_1 + T_3)}2\$\$\implies T =\dfrac { (900+400)}2 = 650 K\$is the Temperature when output of the engines are equal.Let the efficiency of both engines be \$\eta\$. Now considering both engines efficiency are equal. This gives,\$1-\dfrac {T}{T_1}=1-\dfrac {T_3}{T}\\\implies\dfrac {T}{T_1}=\dfrac {T_3}{T}\\\implies T^2=T_1\times T_3\\\implies T^2=800\times400=320000\$\$T=565.68K\$ is the temperature when efficiencies of the both the engines are equal.

Q: An ideal refrigerator has a freezer at a temperature of \$-{13}^{o}C\$. The coefficient of performance of the engine is \$5\$. The temperature of the air (to which heat is rejected) will be:

Given : \$T_H = -13^o\$ C \$\beta = 5\$\$\therefore\$ \$T_H = 273-13 = 260\$ KCoefficient of performance \$\beta = \dfrac{T_H}{T_H-T_L}\$\$\therefore\$ \$5= \dfrac{260}{260-T_L}\$ \$\implies T_L = 312\$ KThus temperature of air \$T_L = 312-273 = 39^oC\$

Question 1

Derive an expression for work done in adiabatic expansion

Accepted Answer

Adiabatic process :    $PV^{\gamma} = K$So,     $P = KV^{-\gamma}$Work done    $W = \int PdV$Or     $W = \int KV^{-\gamma} dV$Or     $W = K \times \dfrac{V^{-\gamma+1}}{1-\gamma}\bigg|_{V_1}^{V_2}$Or     $W = \dfrac{K}{1-\gamma}\times [{V_2^{-\gamma+1}-V_1^{-\gamma+1}}]$Or     $W = \dfrac{1}{1-\gamma}\times [{KV_2^{-\gamma+1}-KV_1^{-\gamma+1}}]$Or     $W = \dfrac{1}{1-\gamma}\times [P_2V_2^{\gamma}{V_2^{-\gamma+1}-P_1V_1^{\gamma}V_1^{-\gamma+1}}]$Or     $W = \dfrac{P_2V_2-P_1V_1}{1-\gamma}$

Question 2

During an adiabatic process, the pressure of a gas is found to the proportional to the cube of its temperature. The ratio of $\displaystyle \frac{C_{p}}{C_{v}}$ for the gas is

Accepted Answer

$P\propto  T^{3}$,$PV=nRT$$P\propto T^{3}$$P\propto  (PV)^{3}$$P^{2}V^{3}=constant$$\displaystyle PV^{\dfrac{3}{2}}=constant$$\displaystyle \gamma =\dfrac{3}{2}$

Question 3

Which of the following statements is incorrect?

Accepted Answer

Critical temperature of the substance is the temperature at a substance's critical point.Above $T_c$ we cannot liquify gas keep pressure constant although on $\uparrow$ pressure we can liquify.

Question 4

A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the &#10;height of each step of the staircase is 15 cm, find the power of the &#10;boy.

Accepted Answer

Given mass of body $(m)=50kg$$h=45\times 15=675cm=6.75m$$t=9s$    $\delta =10m$Potential energy $=mgh$                             $=\left( 50\times 10\times 6.75 \right) 5$                             $=33755$Power of the body $=\dfrac { Potential\quad energy }{ Time } $                                $=\dfrac { 3375 }{ 9 } $                                $=375W$

Question 5

A carnot engine working between 300 K and 600 K has work output of 800 J per cycle. The amount of heat energy supplied to the engine from source per cycle will be:

Accepted Answer

Given : For Carnot engine,$\Rightarrow T_1=600K$       $T_2=300K$$\Rightarrow W=800J$We know,Efficiency $(n)=1-\dfrac{T_2}{T_1}=\dfrac{W}{Q}$$\Rightarrow \dfrac{W}{Q}=1-\dfrac{300}{600}$$\Rightarrow Q=\dfrac{W}{1-\dfrac{1}{2}}$$\Rightarrow \dfrac{800}{\dfrac{1}{2}}$$\Rightarrow Q=1600J$i.e, heat given to system is $1600J.$Hence, the answer is $1600.$

Question 6

An ideal gas heat engine operates in carnot cycle between ${227}^{o}C$ and ${127}^{o}C$. It absorbs $6 \times {10}^{4} cal$ of heat at higher temperature. Amount of heat converted into work, is

Accepted Answer

For a Carnot engine, efficiency $e = 1 - \frac{T_2}{T_1}= 1- \frac{127 + 273}{227 + 273} = 1 - \frac{400}{500} = \frac{1}{5}$Now, $e = \frac{Work \space output}{Heat \space input} = \frac{W}{6 \times 10^{4}}$$W = e \times 6 \times 10^{4} = \frac{1}{5} \times 6 \times 10^{4} = 1.2 \times 10^{4} cal$

Question 7

In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas if the initial internal energy of the gas was 30 J. The final internal energy wilI be:

Accepted Answer

dQ = -20J, dW = -8 J, $U_1 = 30J$By 1st law of thermodynamics,dQ = dU + dW$\Rightarrow $ dU = dQ-dW= -20 + 8 = -12JFinal internal energy, $U_2 = U_1 + dU$= 30-12 = 18 J

Question 8

In a given process on an ideal gas $dW = 0$ and $dQ < 0$, then for the gas :

Accepted Answer

From first law of thermodynamics,$dQ=dU+dW$we have $dQ=dU$               $\left( as\quad dW=0 \right) $But $dQ < 0$$\therefore dU < 0$$N{ C }_{ V }\Delta T< 0$or $\Delta T < 0$Hence, the temperature will decrease.

Question 9

Two Carnot engines A and B are operated in series. The first one A receives heat at 800 K and rejects to a reservoir at temperature T. The second engine B receives the heat rejected by the first engine and in turn rejects to a heat reservoir at 300 K. Calculate the temperature T for the following situations:(a) The outputs of the two engines are equal and(b) The efficiencies of two engines are equal

Accepted Answer

Given:$T_1=800K$$T_3=400K$To find:Temperature T (i) when output of two engines are equal(ii) when efficiencies of two engines are equal.Let the output of both engines be W.Let the engine A take $Q_1$ heat as input at temperature $T_1$ and gives out heat $Q_2$ at temperature $T$ the second engine B receive $Q_2$ as input and give out $Q_3$ at temperature $T_2$ to the sink.Work done by engine, A $W =  Q_1 &#8211; Q_2$Work done by engine, B $W = Q_2 &#8211; Q_3$Thus, $Q_1 &#8211; Q_2 = Q_2 &#8211; Q_3$Dividing both sides by $Q_1$,$1 &#8211; \dfrac {Q_2}{Q_1}  = \dfrac {Q_2}{Q_1} &#8211; \dfrac {Q_3}{Q_1}$$\implies 1-\dfrac T{T_1} = \dfrac {Q_2}{ Q_1}\left (1- \dfrac {Q_3}{ Q_2}\right)$$\implies 1-\dfrac T{T_1} = \dfrac {Q_2}{ Q_1}\left (1- \dfrac {T_3}{ T}\right)$$\implies  1-\dfrac T{T_1} = \dfrac T{T_1} \left(1- \dfrac {T_3} T\right)$$\implies \dfrac  {T_1}T &#8211; 1 = 1 - \dfrac {T_3} T$$\implies \dfrac {T_1}T + \dfrac {T_3} T = 2$$\implies \dfrac 1T (T1 + T3) = 2$$\implies T =\dfrac  {(T_1 + T_3)}2$$\implies T =\dfrac { (900+400)}2 = 650 K$is the Temperature when output of the engines are equal.Let the efficiency of both engines be $\eta$. Now considering both engines efficiency are equal. This gives,$1-\dfrac {T}{T_1}=1-\dfrac {T_3}{T}\\implies\dfrac {T}{T_1}=\dfrac {T_3}{T}\\implies T^2=T_1\times T_3\\implies T^2=800\times400=320000$$T=565.68K$ is the temperature when efficiencies of the both the engines are equal.

Question 10

An ideal refrigerator has a freezer at a temperature of $-{13}^{o}C$. The coefficient of performance of the engine is $5$. The temperature of the air (to which heat is rejected) will be:

Accepted Answer

Given :    $T_H = -13^o$ C               $\beta = 5$$\therefore$    $T_H = 273-13 = 260$ KCoefficient of performance      $\beta = \dfrac{T_H}{T_H-T_L}$$\therefore$       $5= \dfrac{260}{260-T_L}$              $\implies T_L = 312$ KThus temperature of air      $T_L = 312-273  = 39^oC$