If a heat engine and a refrigerator are working between the same two temperatures $T_1$ and $T_2$ $(T_1 > T_2)$,then the ratio of the efficiency of the heat engine to the coefficient of performance of the refrigerator is:

  • A
    $\frac{T_1 - T_2}{T_1 T_2}$
  • B
    $\frac{T_1 + T_2}{T_1 T_2}$
  • C
    $\frac{(T_1 - T_2)^2}{T_1 T_2}$
  • D
    $\frac{(T_1 + T_2)^2}{T_1 T_2}$

Explore More

Similar Questions

The initial state of a certain gas is $(P_i, V_i, T_i)$. It undergoes expansion until its volume becomes $V_f$. Consider the following two cases:
$(a)$ The expansion takes place at constant temperature.
$(b)$ The expansion takes place at constant pressure.
Plot the $P-V$ diagram for each case. In which of the two cases is the work done by the gas more?

Two moles of helium gas are taken over the cycle $ABCDA$, as shown in the $P-T$ diagram. The net work done on the gas in the cycle $ABCDA$ is ...... $R$.

In the figure,a container is shown to have a movable (frictionless) piston on top. The container and the piston are made of perfectly insulating material,allowing no heat transfer between the outside and inside. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow heat transfer. The lower compartment is filled with $2$ moles of an ideal monatomic gas at $700 \ K$,and the upper compartment is filled with $2$ moles of an ideal diatomic gas at $400 \ K$. The heat capacities per mole are: for monatomic gas,$C_v = \frac{3}{2} R, C_p = \frac{5}{2} R$; for diatomic gas,$C_v = \frac{5}{2} R, C_p = \frac{7}{2} R$.
$1.$ Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved,the final temperature of the gases will be:
$(A) 550 \ K$ $(B) 525 \ K$ $(C) 513 \ K$ $(D) 490 \ K$
$2.$ Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then the total work done by the gases until they achieve equilibrium will be:
$(A) 250 \ R$ $(B) 200 \ R$ $(C) 100 \ R$ $(D) -100 \ R$
Give the answer for questions $1$ and $2$.

Two samples $A$ and $B$ of a gas,initially at the same pressure and temperature,are compressed from volume $V$ to $V/2$. Sample $A$ is compressed isothermally,and sample $B$ is compressed adiabatically. The final pressure of $A$ is:

One mole of a gas expands such that its volume $V$ changes with absolute temperature $T$ in accordance with the relation $V = K T^2$,where $K$ is a constant. If the temperature of the gas changes by $60 \text{ K}$,then the work done by the gas is ($R$ is the universal gas constant).

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo