Home
Class 11
PHYSICS
One mole of an ideal gas undergoes a pro...

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

`(7)/(9)(P_(0)V_(0))/(nR)`

Text Solution

Verified by Experts

The correct Answer is:
A
`P=P_(0)[1+(2V_(0)//V)^(2)]^(-1)`
At `V=V_(0),P=P_(0)//5`
`T_(i)=(PV)/(nR)=((P_(0)//5)V_(0))/(nR)=(P_(0)V_(0))/(5n R)`
At `V=2V_(0)P=P_(0)//5`
`T_(i)=(PV)/(nR)=((P_(0)//5)V_(0))/(nR)=(P_(0)V_(0))/(5nR)`
At `V=2V_(0),P=P_(0)//2`
`T_(f)=(PV)/(nR)=((P_(0)//2)(2V_(0)))/(nR)=(P_(0)V_(0))/(5nR)`
`DeltaT=T_(f)-T_(i)=(4P_(0)V_(0))/(5nR)`
Promotional Banner

Topper's Solved these Questions

  • KINETIC THEORY OF GASES

    NARAYNA|Exercise LEVEL-VI|10 Videos

  • KINETIC THEORY OF GASES

    NARAYNA|Exercise LEVEL-I (H.W)|12 Videos

  • KINETIC THEORY OF GASES

    NARAYNA|Exercise NCERT BASED QUESTIONS|28 Videos

  • GRAVITATION

    NARAYNA|Exercise EXERCISE -IV|40 Videos

  • LAW OF MOTION

    NARAYNA|Exercise EXERCISE - IV|35 Videos

Similar Questions

Explore conceptually related problems

One mole of an ideal gas undergoes a process p=(p_(0))/(1+((V_(0))/(V))^(2)) . Here, p_(0) and V_(0) are constants. Change in temperature of the gas when volume is changed from V=V_(0) to V=2V_(0) is

One mole of an ideal gas undergoes a process p=(p_(0))/(1+((V)/(V_(0)))^(2)) where p_(0) and V_(0) are constants. Find temperature of the gaas when V=V_(0) .

One mole of a diatomic gas undergoes a process P = P_(0)//[1 + (V//V_(0)^(3))] where P_(0) and V_(0) are constant. The translational kinetic energy of the gas when V = V_(0) is given by

One mole of diatomic gas undergoes a process P=(P_0)/([1+(V//V_0)^3]) , where P_0 and V_0 are constants . The translational kinetic energy of the gas when V=V_0 is given by

Pressure of an ideal gas is given by P=P_(0)=[1-(1)/(2)((V_(0))/(V))^(2)]

For one mole of ideal gas if P=(P_(0))/(1+((V)/(V_(0)))) where P_(0) and V_(0) are constant, then temperature of gas when V=V_(0) is:

One mole of an ideal gas passes through a process where pressure and volume obey the relation P=P_0 [1-1/2 (V_0/V)^2] Here P_0 and V_0 are constants. Calculate the change in the temperature of the gas if its volume changes from V_0 to 2V_0 .

One mole of an ideal gas undergoes a process P=P_(0)-alphaV^(2) where alpha and P_(0) are positive constant and V is the volume of one mole of gas Q. When temperature is maximum, volume is