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

Thermodynamics Quiz 2

physics

Which of the following statements are true.$\\$ 1. The efficiency of an engine is always < 1.$\\$ 2. Energy is not conserved when heat changes to other forms.$\\$ 3. Energy cannot be transferred from lower temperature to higher temperature by itself. $\\$ 4. All heat energy can be converted to work. 

Analyze the given statements on the basis of $\\$ 1. Efficiency of an engine is given by, $\eta =\dfrac{\text{output}}{\text{input}}$ and there is no perfect engine.$\\$ 2. Energy is always conserved according to the law of conservation of energy.$\\$ 3. Second law of thermodynamics states that Energy cannot be transferred from lower temperature to higher temperature without an external work.$\\$ 4. There will be always come factors acting against a particular work for which energy will be dissipated. 

  1. Efficiency of an engine is given by, $\eta =\dfrac{\text{output}}{\text{input}}=1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$.No engine can convert the given input energy completely into useful work. So, $\eta $ will always be < 1.Only a perfect engine which converts the given input completely into useful work will have the efficiency equal to 1. $\\$ 2. Energy can neither be created nor be destroyed. When heat changes to other forms, there will be conservation of energy.$\\$ 3. According to the second law of thermodynamics, Energy cannot be transferred from lower temperature to higher temperature without an external work. So, Energy cannot be transferred from lower temperature to higher temperature by itself. $\\$ 4. Heat energy cannot be changed completely into useful work as there will be dissipation of energy against the resistances against the work. That means all work cannot be converted to work.Therefore only statements 1 and 3 are correct. 

You are travelling with your family by car. While riding, one of the tires got punctured and air started to leak out quickly through the hole. You tried to locate the puncture by touching on the tube.The puncture hole and its near surroundings will feel ______. 

Adiabatic expansion is the process which plays a major role in inflated tyres. In adiabatic expansion, heat is not transferred between system and surroundings. So, work is done by the gas. 

  $\text{Leaking of air out of a tube is an adiabatic expansion process.}\\ \text{The pressure inside the tyre is higher than that of atmospheric pressure.}\\ \text{So, when the tyre is inflated, air inside the tyre comes in contact with the surroundings and as a result,}\\ \text{internal energy decreases and temperature drops down.}\\ \text{So, air escaping will be cooler than the atmospheric air}\\ \text{so as the hole through which air is leaking out.}\\ \Delta u=\Delta q+W \\ \Delta u=W\text{}\left( \because \Delta q=0 \right) \\ \text{The puncture hole and its near surroundings will feel Colder than other parts of the tube.}$

Assume that a scientist is performing a cyclic process $XYZX$ with an ideal gas by keeping the mass of the gas constant. When he tried to plot a Pressure vs Volume diagram for the same process, he got it as follows<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/266f9424-444a-4aea-bf29-4848b477581d3962810373616672664.png"/><br>Then, which of the following will represent the same process on a $V-T$ graph. 

Analyze each graph given in the option and compare it with the changes happening in the $PV$ diagram for the same process. Arrive at a conclusion after considering each process. Here, as we need to find the $VT$ graph and the given diagram is of $PV$, it will be easier if we analyze them on the basis of change in volume. 

 $\text{If we analyze the given P-V diagram, we can see that, for the process XY,}\\ \text{volume increases as X is near to zero and Y is somewhat further.}\\ \text{Let’s say V. so, for XY, in the V-T graph also, volume must increase.}\\ \text{Now, if we look at the given options,}\\ \text{only in the V-T graph given in option b, volume is increasing.}\\ \text{For rest all graphs, either volume remains constant or the reaction is happening as YX.}$ 

You and your mother are going to a shop to purchase some cooking utensils. In the shop, there are utensils made of $4$ different materials. Which one would you choose $(\text{Don’t go for a cost effective one})$? 

As we are considering the given 4 materials irrespective of their cost, the most preferable material for cooking utensils is the one having least specific heat and highest conductivity. 

 For better cooking, the material from which the utensil was made from must be having low specific heat and high conductivity.Specific heat is the amount of heat energy needed to be supplied for increasing the temperature by $1{}^\circ C$ . so, if specific heat is low, we will only need to supply low amount of heat energy. Higher conductivity helps to give maximum heat to the thing which we are cooking.Otherwise, the food will take more time to get properly cooked. Here, the order of specific heat is like, Lead < Copper < Stainless Steel < Aluminum. And order of conductivity is like, Stainless Steel < Lead < Aluminum < Copper. Therefore, copper is the best material for cooking utensils. 

You are provided with a relation for an ideal gas which is undergoing adiabatic reaction as $U=k+3PV$. Where $k$ is a constant. Then the type of gas in the process will be, 

 Try to change the given relation into the form of a standard adiabatic reaction equation using differentiation and integration. Compare those equations and find $\gamma $ and hence the type of gas. 

 $\text{The relation given is, U=k+3PV. By differentiating both sides, we get}\\ dU=0+3(PdV+VdP)\text{   -- (1)}\\ \text{We must know that, for an adiabatic change,}\\ dQ=0=dU+dW=dU+PdV \\  \Rightarrow dU=-PdV\\ \text{Substitute this in the (1). We will get,}\\ -PdV=3(PdV+VdP)\text{ } \\  \Rightarrow 4PdV=-3VdP \\  \Rightarrow 4\left( \dfrac{dV}{V} \right)=-3\left( \dfrac{dP}{P} \right) \\ \text{ Now, integrating both sides,}\\ \Rightarrow 4\log V+3\log P=\text{constant} \\  \Rightarrow \log {{P}^{3}}{{V}^{4}}=\text{constant} \\  \Rightarrow P{{V}^{\dfrac{4}{3}}}=\text{constant}\\ \text{ Compare this equation with the standard equation of adiabatic change. Which is,}\\ P{{V}^{\gamma }}=\text{constant}\\ \text{ Then we get,}\, \gamma =\dfrac{4}{3}.\\ \text{Therefore the gas must be a polyatomic gas.}$