Home
Class 12
PHYSICS
n mole a perfect gas undergoes a cyclic ...

n mole a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.
`A to B`: Isothermal expansion at temperature T so that the volume is doubled from `V_1` to `V_2 = 2V_1` and pressure changes from `P_1` to `P_2`
`B to C` : Isobaric compression at pressure `P_2` to initial volume `V_1`
` C to A` : Isochoric change leading to change of pressure from `P_2` to` P_1` .
Total work done in the complete cycle ABCA is :

A

0

B

`nRT (ln2 + 1/2)`

C

`nRTln2`

D

`nRT (ln2 - 1/2)`

Text Solution

Verified by Experts

The correct Answer is:
D
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • JEE MAIN 2021

    JEE MAINS PREVIOUS YEAR|Exercise SECTION-B|40 Videos

  • JEE MAIN 2021

    JEE MAINS PREVIOUS YEAR|Exercise PHYSICS (SECTION-A)|20 Videos

  • JEE MAIN 2021

    JEE MAINS PREVIOUS YEAR|Exercise PHYSICS SECTION B|30 Videos

  • JEE MAIN

    JEE MAINS PREVIOUS YEAR|Exercise All Questions|473 Videos

  • JEE MAIN 2022

    JEE MAINS PREVIOUS YEAR|Exercise Question|492 Videos

Similar Questions

Explore conceptually related problems

One mole of an ideal gas goes from an initial state A to final state B via two processes : It first undergoes isothermal expansion from volume V to 3V then its volume is reduced from 3V to V at constant pressure. The correct P-V diagram representing the two processes is

One mole of an ideal gas goes from an initial state A to final state B via two processes . It first undergoes isothermal expansion from volume V to 3V and then its volume is reduced from 3 V to V at constant pressure . The correct P - V diagram representing the two processes is

Knowledge Check

  • A cyclic process ABCA is shown in the V-T diagram process on the P-V

    A
    B
    C
    D

  • A gas expands under constant pressure P from volume V_(1) to V_(2) The work done by the gas is

    A
    `P(V_(2)-V_(1))`
    B
    `P(V_(1)-V_(2))`
    C
    `P(V_(1)^(lambda)-V_(2)^(lambda))`
    D
    `P(V_(1)V_(2))/(V_(2)-V_(1))`

  • One mole of an ideal gas undergoes a process P = P_(0) [1 + ((2 V_(0))/(V))^(2)]^(-1) , where P_(0) V_(0) are constants. Change in temperature of the gas when volume is changed from V = V_(0) to V = 2 V_(0) is:

    A
    `(4)/(5) (P_(0) V_(0))/(nR)`
    B
    `(3)/(4) (P_(0) V_(0))/(nR)`
    C
    `(2)/(3) (P_(0) V_(0))/(nR)`
    D
    `(9)/(7) (P_(0) V_(0))/(nR)`

    • Similar Questions

    Explore conceptually related problems

    One mole of an ideal gas goes from an initial state A to final state B via two processs : It first undergoes isothermal expansion from volume V to 3 V and then its volume is reduced from 3 V to V at constant pressure. The correct P-V diagram representing the two process in (figure)

    One mole of ideal gas goes through process P = (2V^2)/(1+V^2) . Then change in temperature of gas when volume changes from V = 1m^2 to 2m^2 is :

    One mole of ideal gas gose through process P=(2V^(2))/((1+V^(2))) then change in temperature of the gas when volume changes from V=1m^(3) to 2m^(3) is :

    One mole of an ideal monotomic gas undergoes a linear process from A to B in which its pressure p and its volume V change as shown in figure. The absolute temperature T versus volume V for the given process is

    One mole of an ideal monatomic gas undergoes a linear process from A to B , in which is pressure P and its volume V change as shown in figure . The absolute temperature T versus volume V for the given process is

    Home
    Profile