A Carnot engine operating between temperatures T₁ and T₂ has efficiency 0.2. When T₂ is reduced by 50 K, its efficiency increases to 0.4. Then, T₁ and T₂ are respectively

• 200 K, 150 K

• 250 K, 200 K

• 300 K, 250 K

• 300 K, 200 K

If the temperatures of source and sink of a Carnot engine having efficiency η are each decreased by 100 K, then the efficiency

• remains constant

• becomes 1

• decreases

• increases

The p-V diagram of a gas system undergoing cyclic process is shown here. The work done during isobaric compression is

• 100 J

• 200 J

• 400 J

• 500 J

34.

In a cyclic process, the amount of heat given to a system is equal to

• net increase in internal energy

• net work done by the system

• net decrease in internal energy

• net change in volume

Identify the wrong statement

• For isothermal process, ΔT = 0

• For isochoric process, ΔV = 0

• For cyclic process, ΔW = 0

• For adiabatic process, ΔQ = 0

A Carnot engine whose efficiency is 40%, receives heat at 500 K. If the efficiency is to be 50%, the source temperature for the same exhaust temperature is

• 900 K

• 600 K

• 700 K

• 800 K

The thermodynamic process in which no work is done on or by the gas is

• isothermal process

• adiabatic process

• isochoric process

• isobaric process

A refrigerator with coefficient of performance $\frac{1}{3}$ releases 200 J of heat to a hot reservoir. Then the work done on the working substance is

• $\frac{100}{3} \mathrm{J}$

• 100 J

• $\frac{200}{3} \mathrm{J}$

• 150 J

The heat capacity per mole of water is (R is universal gas constant)

• 9 R

• $\frac{9}{2} \mathrm{R}$

• 6 R

• 5 R

The change in internal energy of a given mass of gas, when its volume changes from V to 2V at constant pressure p is $\left(\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}} = \mathrm{\gamma }, \mathrm{universal} \mathrm{gas} \mathrm{constant} = \mathrm{R}\right)$

• $\frac{\mathrm{pV}}{\mathrm{\gamma }}$

• $\frac{\mathrm{pV}}{\left(2\mathrm{\gamma } - 1\right)}$

• $\frac{\mathrm{pV}}{\left(\mathrm{\gamma } -1\right)}$

• $\frac{\mathrm{R}}{\left(\mathrm{\gamma } - 1\right)}$