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This is second of our many quizzes on Thermal energy. Click on this to access more such competitive quizzes on Thermal Physics.

Thermal energy Quiz 2

physics

In an isobaric process if volume is increased by $10\,{{m}^{3}}$, then the temperature increases by $20\%$. What was the initial volume?

As pressure is constant use $P=$constant in the ideal gas equation $PV=nRT$. Process with a constant pressure is also known as an isobaric process. 

  $As\text{ }P=\text{constant},\text{ }n=\text{constant} \\PV=nRT \\ \Rightarrow V\propto T \\ \Rightarrow\dfrac{\Delta V}{{{V}_{i}}}=\dfrac{\Delta T}{{{T}_{i}}} \\ \Rightarrow\dfrac{\Delta V}{{{V}_{i}}}\times 100=\dfrac{\Delta T}{{{T}_{i}}}\times 100 \\ \text{as  }\!\!\%\!\!\text{  change in temperature is }20\%, \\ \dfrac{\Delta T}{{{T}_{i}}}\times 100=20 \\ \Rightarrow \dfrac{\Delta V}{{{V}_{i}}}\times 100=20 \\ \text{Given }\Delta V=10\text{ }{{\text{m}}^{\text{3}}} \\{{V}_{i}}=\dfrac{10\times 100}{20} \\ \therefore{{V}_{i}}=50\text{ }{{m}^{3}}$

 If an ideal gas is filled in container with the same number of moles as the ideal gas, then $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4dd30391-ed00-45b4-8c2b-aa375966b5d43145980962033742838.png"/> 

To compare temperature in $P-V$ diagram use $P=$ constant line or $V=$ constant line and compare results. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b341ade7-6b87-4b43-aeda-96abbbe1369e5146970630498696780.png"/>$ \\ \text{For an ideal gas, }PV=nRT \\ \text{In given graph if we draw a constant pressure line at }{{\text{P}}_{{\mathrm O}}}\text{.} \\ \text{Then for }P={{P}_{{\mathrm O}}} \\ \therefore V\propto T \\ \text{Then from graph }{{\text{V}}_{B}}>{{V}_{A}} \\Thus,\,\,{{T}_{B}}>{{T}_{A}} \\ \text{Similarly, if we draw a constant volume line at }V={{V}_{{\mathrm O}}},\text{then} \\P\propto T \\ \text{Then again from the graph, }{{P}_{B}}>{{P}_{A}} \\ \therefore \,\,{{T}_{B}}>{{T}_{A}} \\Thus\text{ }{{T}_{B}}>{{T}_{A}},\,\,always.$ 

 The region under the closed curve in the $T-S$ diagram denotes $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5cbdf238-4976-43be-8342-2b6d1711f17f7779885356273987623.png"/> 

Area under $T-S$ diagram gives value of heat interaction. Where $T$ is temperature and $S$ is entropy of fluid. 

 $\text{Area under the }T-S\text{ diagram gives} \\ \text{Heat involvement in a thermodynamic process as} \\A=\Delta T\Delta S \\And\text{ }\Delta S=\dfrac{\Delta {{Q}_{net}}}{\Delta T}\text{ is the definition of entropy} \\ \therefore A=\Delta {{Q}_{net}} \\ \text{Thus, it }\!\!'\!\!\text{ s the net heat energy involved}\text{.}$

If length of arc $AC=3.14\,m$ then find net work done in the complete cycle for the $P-V$ diagram.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c0a2b42b-bc48-4169-8154-b46db93b98e16505533843861476463.png"/>

To calculate work done find out the area under $P-V$ diagram. $W=PV$, Where $P$ denotes pressure and $V$ denotes fluid volume.

 $Length\text{ of Arc AC= 3}\text{.14}\,metre \\Length\,\,of\,\,AC\Rightarrow \dfrac{{{90}^{{\mathrm O}}}}{{{360}^{{\mathrm O}}}}\times 2\pi r=\text{3}\text{.14} \\ \Rightarrow r=\dfrac{2\times 3.14}{3.14} \\ \Rightarrow r=2\,\,metre \\ \text{Area under }P-V\text{ diagram gives work done in process} \\ \text{Area under closed loop gives net work done in a complete cycle} \\Area\,\,under\,\,circle\,\,A=\pi {{r}^{2}}=\pi \times 2\times 2 \\ \therefore A =4\pi \,\,Joules$

 Which of the two curves is most likely to be adiabatic?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/417e32db-7ec7-4414-87b1-e7dfcd52c5f15280903017777751144.png"/> 

To predict the type of process use $P{{V}^{n}}=nRT$ and compare the value of  $''n''$ for different processes. 

 $For\text{ Isothermal process, T=constant} \\ \text{PV=Constant} \\ \text{differentiating with respect to V}, \\ \dfrac{dP}{dV}\times V+P=0 \\{{\left. \dfrac{dP}{dV} \right|}_{isothermal}}=\dfrac{-P}{V} \\For\text{ Adiabatic process,} \\ \text{P}{{\text{V}}^{\gamma }}\text{=Constant} \\ \text{differentiating with respect to V,} \\ \dfrac{dP}{dV}\times {{V}^{\gamma }}+\gamma P{{V}^{\gamma -1}}=0 \\ \Rightarrow {{\left. \dfrac{dP}{dV} \right|}_{adiabatic}}=\dfrac{-\gamma P}{V} \\Neglecting\text{ the negative sign and comparing magnitudes of the slopes,} \\{{\left. \dfrac{dP}{dV} \right|}_{isothermal}}<{{\left. \dfrac{dP}{dV} \right|}_{adiabatic}} \\ \text{Thus, the curve for an adiabatic process is steeper than isothermal process}\text{.} \\ \text{So 2 is an adiabatic process}\text{.}$