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Thermal processes Quiz 2

physics

 The P-V diagram of an ideal gas undergoing a certain thermal process is given below, determining the change in internal energy of the gas. Consider specific heat capacity of the gas to be $24\text{ J}\cdot \text{mo}{{\text{l}}^{-1}}\cdot {{\text{K}}^{-1}}$<br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5c510917-b4ba-4351-b571-84110039449c1797050586687326545.png"/>

 The area under the P-V curve depicting the process on the P-V plane is the work done during that process. 

  $\text{Work done for any process is given by;} \\ W=\text{ Area under curve} \\ \Rightarrow W=\dfrac{1}{2}\left( 16-4 \right)\left( 8-2 \right)+12\times 2 \\ \Rightarrow W=3\times 12+2\times 12 \\ \Rightarrow W=36+24 \\ \Rightarrow W=60\text{ J} \\ \text{Change in heat is given by;} \\ \Delta Q=nC\Delta T \\ \Rightarrow \Delta Q=24\times 100 \\ \Rightarrow \Delta Q=2400\text{ J} \\ \text{From the first law of thermodynamics;} \\ \Delta U=\Delta Q-W=2400-60 \\ \therefore \Delta U=2440\text{ J}$

 An ideal gas is undergoing an adiabatic compression. If pressure changes from $P$ to $4P$ and volume decreases from $2V$ to $V$ , then what is the nature of the ideal gas if work done is $5PV$?

 The work done in case of an adiabatic process is inversely proportional to $\gamma -1$ . $\gamma $ depends on the nature of gas. 

  $\text{For adiabatic process;} \\ W=\dfrac{{{P}_{1}}{{V}_{1}}-{{P}_{2}}{{V}_{2}}}{\gamma -1} \\ \text{Given that }W=5PV \\ \Rightarrow W=-5PV\text{                    }\left( \because \text{ The gas is compressed} \right) \\ \Rightarrow -5PV=\dfrac{2PV-4PV}{\gamma -1} \\ \Rightarrow \gamma -1=\dfrac{2PV}{5PV}=\dfrac{2}{5} \\ \Rightarrow \gamma =\dfrac{2}{5}+1 \\ \therefore \gamma =\dfrac{7}{5} \\ \gamma =\dfrac{7}{5}\text{ is for diatomic gas} \\ \text{Therefore, the given gas is a diatomic gas}$

 Evaluate the ${{C}_{p}}$ and ${{C}_{v}}$ for an ideal gas undergoing adiabatic expansion if volume of the gas changes from $V$ to $4V$ and temperature changes from $5T$ to $T$ . Slope of the V-T graph is $-\sqrt{3}$. 

 Using the volume-temperature relation for an adiabatic process, determine the change in volume with respect to temperature. 

 $\text{For adiabatic process;} \\ T{{V}^{\gamma -1}}=\text{ Constant} \\ \text{Differentiating the above equation;} \\ T\left( \gamma -1 \right){{V}^{\gamma -2}}dV+{{V}^{\gamma -1}}dT=0 \\ \Rightarrow T\left( \gamma -1 \right)\dfrac{{{V}^{\gamma -1}}}{V}dV+{{V}^{\gamma -1}}dT=0 \\ \text{Dividing both sides by }{{V}^{\gamma -1}} \\ \Rightarrow \dfrac{T}{V}\left( \gamma -1 \right)dV=-dT \\ \Rightarrow \dfrac{dV}{dT}=\dfrac{-V}{T\left( \gamma -1 \right)} \\ \text{Given that;} \\ \dfrac{dV}{dT}=-\sqrt{3} \\ \Rightarrow \dfrac{-V}{T\left( \gamma -1 \right)}=-\sqrt{3} \\ \Rightarrow \dfrac{V}{T\sqrt{3}}=\gamma -1 \\ \Rightarrow \gamma =\dfrac{V+T\sqrt{3}}{T\sqrt{3}} \\ \text{Specific heat at constant volume is given by;} \\ \therefore {{C}_{v}}=\dfrac{R}{\gamma -1}=\dfrac{TR\sqrt{3}}{V} \\ \text{Specific heat due to constant pressure is given by;} \\ \therefore {{C}_{p}}=\dfrac{\gamma R}{\gamma -1}=R\left( V+T\sqrt{3} \right)$

 Four moles of an ideal gas undergo a polytropic process. If the work done is $-218.88\text{ J}$, what type of gas is used and also what conclusions do you draw about the process followed?

 A polytropic process is a process where the equation of pressure versus volume curve is given by $P{{V}^{N}}=\text{ Constant}$. 

 $\text{For polytropic process;} \\ W=\dfrac{-nR\Delta T}{N-1} \\ \text{Given that;} \\ W=-218.88\text{ J} \\ \Rightarrow -218.88=\dfrac{-4\times 0.0821\times 1000}{N-1} \\ \Rightarrow N-1=\dfrac{4\times 82.1}{218.88}=1.5 \\ \therefore N=2.5=\dfrac{5}{2} \\ \text{But }\gamma =\dfrac{5}{2}\text{ for monoatomic gas} \\ \therefore N=\gamma =\dfrac{5}{2} \\ \text{Therefore, the process is an adiabatic process and gas is a monatomic gas}\text{.}$

 $n$ moles of an ideal gas undergo a cyclic process $ABCDA$ , such that $AB$ and $CD$ are isobaric processes with pressure for processes $AB$ and $CD$ as $2p$ and $p$ respectively, $BC$ and $DA$ are isochoric processes with volumes for the processes $BC$ and $DA$ as $2v$ and $v$ respectively. Determine the efficiency of the cycle. Take the values of ${{C}_{p}}$ and ${{C}_{v}}$ as $\dfrac{5}{2}R$ and $\dfrac{3}{2}R$ respectively. 

 Efficiency of a thermodynamic cycle is defined as the ratio of the net work done to the total heat supplied. 

 &nbsp;$\text{Net work done is given by;} \\ \Rightarrow W=\left( 2p-p \right)\left( 2v-v \right) \\ \Rightarrow W=pv \\ \text{Heat supplied for process DA is given by;} \\ \Rightarrow \Delta {{Q}_{1}}=\Delta U+W \\ W=0\text{ for isochoric} \\ \Rightarrow \Delta {{Q}_{1}}=\Delta U=n{{C}_{v}}\Delta T \\ \Rightarrow \Delta T={{T}_{1}}-{{T}_{2}}=\dfrac{2pv}{nR}-\dfrac{pv}{nR}=\dfrac{pv}{nR} \\ \text{Given that }{{C}_{v}}=\dfrac{3}{2}R; \\ \Rightarrow \Delta {{Q}_{1}}=n\left( \dfrac{3}{2}R \right)\left( \dfrac{pv}{nR} \right)=\dfrac{3pv}{2} \\ \text{Heat supplied during process AB is given by;} \\ \text{Process AB is an isobaric process} \\ \Rightarrow \Delta {{Q}_{2}}=n{{C}_{p}}\Delta T \\ \Rightarrow \Delta T=\dfrac{4pv}{nR}-\dfrac{2pv}{nR}=\dfrac{2pv}{nR} \\ \text{Given that }{{C}_{p}}=\dfrac{5}{2}R; \\ \Rightarrow \Delta {{Q}_{2}}=n\left( \dfrac{5}{2}R \right)\left( \dfrac{2pv}{nR} \right)=5pv \\ \Rightarrow \text{Total heat supplied = }\Delta {{Q}_{total}}=\Delta {{Q}_{1}}+\Delta {{Q}_{2}}=\dfrac{13pv}{2} \\ \text{Efficiency of a cycle is given by;} \\ \eta =\dfrac{\text{net work done}}{\text{total heat supplied}}=\dfrac{2pv}{13pv}=0.1538 \\ \therefore {{\eta }_{\text{percent}}}=0.1538\times 100=15.38\%$