Monatomic, diatomic and polyatomic ideal gases each undergo slow adiabatic expansions from the same initial volume and same initial pressure to the same final volume.The magnitude of the work done by the environment on the gas is

the greatest for the polyatomic gas
the greatest for the monatomic gas
the greatest for the diatomic gas
the question is irrelevant, there is no meaning of slow adiabatic expansion

Solution

\(W=\frac{nRdT}{\gamma -1}\) γis minimum for a polyatomic gas Hence, W is greatest for polyatomic gas

During an adiabatic process an object does 100J of work and its temperature decreases by 5K. During another process it does 25J of work and its temperature decreases by 5K. Its heat capacity for 2ndprocess is

20 J/K
24 J/K
15 J/K
100 J/K

Solution

For adiabatic process, dU = – 100 J

which remains same for other processes also.

Let C be the heat capacity of 2ndprocess then

– (C)5= dU + dW

= – 100 + 25 = – 75

∴ C = 15 J/K

Volume of one mole gas changes according to the V = a/T. If temperature change is ΔT, then work done will be

RΔT
-RΔT
\(\frac{R}{\gamma -1}\Delta T\)
R (γ– 1)ΔT

Solution

For an ideal gas graph is shown for three processes. Process1, 2 and 3 are respectively.

Isobaric, adiabatic, isochoric
Adiabatic, isobaric, isochoric
Isochoric, adiabatic, isobaric
Isochoric, isobaric, adiabatic

Solution

Isochoric proceess dV = 0

W = 0 proceess 1

Isobaric : W = PΔV = nRΔT

Adiabatic |W |= \(\frac{nR\Delta T}{\gamma -1}\) 0< γ– 1< 1 As work done in case of adiabatic process is more soprocess 3 is adiabatic and process 2 is isobaric.

A Carnot engine works first between 200°C and 0°C and then between 0°C and –200°C. The ratio of its efficiency in the two cases is

1.0
0.577
0.34
0.68

Solution

Two cylinders fitted with pistons contain equal amount of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of 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 30 K, then the rise in temperature of gas in B is

30 K
18 K
50 K
42 K

Solution

In pressure-volume diagram, the is ochoric, is othermal,isobaric and is o-entropic parts respectively, are

BA, AD, DC,CB
DC, CB, BA, AD
AB, BC, CD, DA
CD, DA, AB, BC

Solution

From C to D, V is constant. So process is isochoric.From D to A, the curve represents constant temperature. So the process is isothermal.From A to B, pressure is constant . So,the process isisobaric.BC represents constant entropy.

A uniform sphere is supplied heat electrically at the centre at a constant rate. In the steady state, steady temperatures are established at all radial locations r,heat flows out war dsradial and is ultimately radiated out by the outer surface isotropically. In this steady state, the temperature gradient varies with radial distance r according to

r^–1
r^–2
r^–3
r^–3/2

Solution

Flow rate α gradient × r².

When flow rate is constant, gradient α r^-2.

Ice contained in a beaker starts melting when

the specific heat of the system is zero
internal energy of the system remains constant
temperature remains constant
entropy remains constant

Solution

During melting temperature remains constant

One mole of anideal gas at temperature T was cooled isochorically till the gas pressure fell from P to ^p⁄_n. Then,by an isobaric process, the gas was restored to the initial temperature. The net amount of heat absorbed by the gas in the process is

nRT
^RT⁄_n
RT (1– n^–1)
RT (n – 1)

Solution

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete