Home
Class 11
CHEMISTRY
Initially one mole of ideal gas (C(v)=(5...

Initially one mole of ideal gas `(C_(v)=(5)/(2)R)` at `0.1` atm and 300 K is put through the following cycle:

Step-I : Heating to twice its initial pressure at constant volume.
Step-II : Adiabatic expansion to its initial temperature.
Step-III : Isothermal compression back to `1.00` atm.
what is the volume at state X?

A

`40.4` L

B

`65.0` L

C

139 L

D

4.35 L

Text Solution

Verified by Experts

The correct Answer is:
C
`V_(1)=24.63 "L " T_(1)=300 K , T_(2)=600 K`

For path (II)
`T_(1)V_(1)^(gamma-1)=T_(2)V_(2)^(gamma-1)`
`(T_(1))/(T_(2))=((V_(2))/(V_(1)))^(gamma-1)`
`(600)/(300)=((V_(2))/(V_(1)))^((7)/(5)-1)`
`(2)^(5//2)=(V_(2))/(V_(1))`
`V_(2)=4sqrt(2)xx24.63 = 139.3 L~~ 139 L`
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    NARENDRA AWASTHI|Exercise Level 1 (Q.1 To Q.30)|6 Videos

  • THERMODYNAMICS

    NARENDRA AWASTHI|Exercise Level 1 (Q.31 To Q.60)|2 Videos

  • STOICHIOMETRY

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|20 Videos

Similar Questions

Explore conceptually related problems

One mole of an ideal gas at an initial temperature of TK does 6R joules of work adiabatically. If the ratio of specific hats of this gas a constant pressure and at constant volume is 5/3, the final temperature (T_f) of gas will be :

An ideal gas has volume V_0 at 27^@C . It is heated at constant pressure so that its volume becomes 2V_0 . The final temperature is

At a constant temperature a gas is initially at 2 atm pressure. To compress it to 1/8th of its initial volume, pressure to be applied is

When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas (Delta S) is

An ideal gas is found to obey pV^(3/2) = constant during an adiabatic process. If such a gas initially at a temperature T is adiabatically compressed to half of its initial volume, then its final temperature is