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Important Questions on Thermodynamics

HARD
The heat capacity of one mole an ideal is found to be CV=3R1+aRT2 where a is constant. The equation obeyed by this gas during a reversible adiabatic expansion is:
HARD
Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u=UVT4 and pressure p=13UV . If the shell now undergoes an adiabatic expansion the relation between T and R is:
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One mole of an ideal monatomic gas undergoes a process described by the equation PV3= constant. The heat capacity of the gas during this process is
MEDIUM
The bulk modulus of a gas is defined as, B=-VdPdV . For an adiabatic process, the variation of B is proportional to Pn. For an ideal gas, n is
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One mole of an ideal monatomic gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100K and the universal gas constant R=8.0 J mol-1 K-1, then how much is the decrease in its internal energy (in J) ?
HARD
 An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is:

(Take Cv=1.5R, whereR is gas constant)

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HARD
One mole of an ideal monatomic gas is taken along the path ABCA as shown in the P-V diagram. The maximum temperature attained by the gas along the path BC is given by:
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One mole of a monoatomic ideal gas expanded by a process described by PV3=C where C is a constant. The heat capacity of the gas during the process is given by ( R is the gas constant)
MEDIUM
An ideal gas goes through a reversible cycle abcd has the V - T diagram shown below. Process da and bc are adiabatic.

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The corresponding P - V diagram for the process is (all figures are schematic and not drawn to scale) :
HARD
A gas obeying the equation of state PV=RT undergoes a hypothetical reversible process described by the equation, PV53exp-PVE0=C1, where C1 and E0 are dimensioned constants. Then, for this process, the thermal compressibility at high temperature
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Thermodynamic processes are indicated in the following diagram.

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Match the following

Column – 1 Column - 2
P. Process I a. Adiabatic
Q. Process II b. Isobaric
R. Process III c. Isochoric
S. Process IV d. Isothermal
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Half mole of an ideal monoatomic gas is heated at a constant pressure of 1atm from 20° C to 90° C. Work done by the gas is(Gas constant,R=8.21 J mol-1 K-1)
MEDIUM

In the P-V diagram below the dashed curved line is an adiabatic

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For a process that is described by a straight line joining two points X and Y on the adiabatic (solid line in the diagram), heat is: (Hint: Consider the variations in temperature from X and Y along the straight line)

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A gas at initial temperature T undergoes sudden expansion from volume V to 2V. Then.
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The three processes in a thermodynamic cycle shown in the figure are : Process 12 is isothermal; Process 23 is isochoric (volume remains constant); Process 13 is adiabatic.

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The total work done by the ideal gas in this cycle is,10 J. The internal energy decreases by, 20 J in the isochoric process. The work done by the gas in the adiabatic process is, -20 J. The heat added to the system in the isothermal process is

MEDIUM

Two moles of an ideal monoatomic gas occupies a volume V at 27oC . The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

MEDIUM
A monoatomic gas is compressed from a volume of 2 m3 to a volume of 1 m3 at a constant pressure of 100 N m2. Then it is heated at constant volume by supplying 150 J of energy. As a result, the internal energy of the gas
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One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,

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The change in internal energy of the gas during the transition is:

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For the given cyclic process CAB as shown for a gas, the work done is:

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A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then: