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One mole of a monoatomic gas is taken from P to R, via three paths PQR, PR and PSR. If work done by the gas in PQR is W1, in PR work done is W2 and in PSR work done is W3, then

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Important Questions on Thermodynamics

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A thermodynamic system undergoes a cyclic process ABC as shown in the diagram. The work done by the system per cycle is 

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The equation of state for a gas is given by PV = nRT + α V , where n is the number of moles and α   is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are T0 and P0 respectively. The work done by the gas when its temperature doubles isobarically will be :
HARD
Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamics process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by TΔX, where T is temperature of the system and ΔX is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas X=32RlnTTA+RlnVVA. Here, R is gas constant, V is volume of gas, TA and VA are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.
 
  Column - I   Column - II
(I) Work done by the system in process 123 (P) 13RT0ln2
(II) Change in internal energy in process 123 (Q) 13RT0
(III) Heat absorbed by the system in process 123 (R) RT0
(IV) Heat absorbed by the system in process 12 (S) 43RT0
    (T) 13RT03+ln2
    (U) 56RT0

If the process on one mole of monatomic ideal gas is as shown in the PV -diagram with P0V0=13RT0, the correct match is

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A gas at state A changes to state B through path I and II shown in figure. The change in internal energy is ΔU1 and ΔU2, respectively. Then

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An ideal monoatomic gas occupies a volume of 2 m3 at a pressure of 3×106 Pa. The energy of the gas is:
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The correct option for free expansion of an ideal gas under adiabatic condition is
HARD
Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamics process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by TΔX, where T is temperature of the system and ΔX is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas X=32RlnTTA+RlnVVA. Here, R is gas constant, V is volume of gas, TA and VA are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.
 
  List I   List II
(I) Work done by the system in process 123 (P) 13RT0ln2
(II) Change in internal energy in process 123 (Q) 13RT0
(III) Heat absorbed by the system in process 123 (R) RT0
(IV) Heat absorbed by the system in process 12 (S) 43RT0
    (T) 13RT03+ln2
    (U) 56RT0

If the process carried out on one mole of monatomic ideal gas is as shown in figure in the T - V -diagram with P0V0=13RT0, the correct match is,

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A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x×10-1 atm. Value of x (nearest integer) is              
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2 kg of a monoatomic gas is at a pressure of 4×104 N m-2. The density of the gas is 8kg m-3. What is the order of energy of the gas due to its thermal motion?
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N moles of diatomic gas in a cylinder is at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. The change in the total kinetic energy of the gas is
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The internal energy U, pressure P and volume V of an ideal gas are related as U =3PV+4. The gas is
EASY
An ideal gas is compressed to half of its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?
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The volume V of a given mass of monoatomic gas changes with temperature T according to the relation V=KT23. The workdone when temperature changes by 90 K will be xR. The value of x is
[R universal gas constant ]
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Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume. The work done is :
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A monoatomic gas of mass 4.0 u is kept in an insulated container. The container is moving with velocity 30 m s-1. If the container is suddenly stopped then a change in temperature of the gas (R=gas constant) is x3R. Value of x is,
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A gas mixture contains monoatomic and diatomic molecules of 2 moles each. The mixture has a total internal energy of (symbols have usual meanings)
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An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is S1 and that of the other part is S2. Given that S1>S2. If the piston is removed then the total entropy of the system will be:
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A gas mixture consists of 2 moles of O2 and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is ( R is universal gas constant)
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An ideal gas is taken reversibly around the cycle a-b-c-d-a as shown on the T (temperature) - S (entropy) diagram

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The most appropriate representation of above cycle on a U (internal energy)-V (volume) diagram is

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One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement :

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