InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
For free expansion of the gas, which of the following is true?A. `Q=W=0` and `DeltaE_("int")=0`B. `Q=0` and `W gt 0` and `DeltaE_("int")=-W`C. `W = 0,Q gt 0` and `DeltaE_("int")=Q`D. `W gt 0`,`Q lt 0` and `DeltaE_("int")=0` |
| Answer» Correct Answer - A | |
| 152. |
The state of a thermodynamic system is represented byA. Pressure onlyB. Volume onlyC. Pressure, volume and temperatureD. Number of moles |
| Answer» Correct Answer - C | |
| 153. |
Assertion: Between two thermodynamic states, the value of (Q-W) is constant for any process. Reason: Q and W are path functions.A. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B `Q-W=DeltaU` is state function. So, it will remain constant for any two thermodynamic states. |
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| 154. |
The given figure shows the variation of force applied by ideal gas on a piston which undergoes a process during which piston position changes from 0.1 to 0.4m. If the internal energy of the system at the end of the process is 2.5 J higher, then the heat absorbed during the process is A. (a) 15JB. (b) 17.5JC. (c) 20JD. (d) 22.5J |
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Answer» Correct Answer - C W= Area under F-x graph `=17.5J` Now, `Q=W+DeltaU=20J` |
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| 155. |
Assertion: First law of thermodynamics can be applied for ideal gases only. Reason: First law is simply, law of consevation of energy.A. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D First law is just energy conservation law, which can be applied for any system. |
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| 156. |
Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K. The piston of A is free to move, while that B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30K, then the rise in temperature of the gas in B isA. (a) 30KB. (b) 18KC. (c) 50KD. (d) 42K |
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Answer» Correct Answer - D `Q_A=Q_B` `nC_pDeltaT_A=nC_VDeltaT_B` `:.` `DeltaT_B=(C_p)/(C_V)*DeltaT_A=gammaDeltaT_A` `=(1.4)(30)=42K` |
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| 157. |
An ideal gas is subjected to cyclic process involving four thermodynamics thates, the amounts of heat `(Q)` and work `(W)` involved in each of these states. `Q_1=6000J,Q_2=-5500J,Q_3=-3000J,Q_4=3500J` `W_1=2500J,W_2=-1000J,W_3=-1200J,W_4=xJ` The ratio of the net work done by the gas to the total heat absorbed by the gas is `eta`. The values of x and `eta` respectively areA. `500,7.5%`B. `700,10.5%`C. `1000,21%`D. `1500,15%` |
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Answer» Correct Answer - B In cyclic process `Q_1+Q_2+Q_3+Q_4=W_1+W_2+W_3+W_4` `6000-5500-3000+3500=2500-1000-1200+x` `x=700J` `DeltaQ_(cyclic)=DeltaW_(cyclic)=1000J` `eta=(DeltaW_("cyclic"))/("Heat supplied")=(1000)/(Q_1+Q_4)=(1000)/(6000+3500)` `=(10)/(95)=(2)/(19)` i.e., `(200)/(19)=10.5%` |
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