Home
Class 11
CHEMISTRY
When one mole of an ideal gas is compres...

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

A

`C_(p,m)ln2`

B

`C_(v,m)ln2`

C

`R ln 2`

D

`(C_(v,m)-R)ln2`

Text Solution

Verified by Experts

The correct Answer is:
D
`DeltaS=nC_(v,m)ln.(T_(2))/(T_(1))+nRln.(V_(2))/(V_(1))`
`=C_(v,m)ln2+Rln((1)/(2))`
`DeltaS=(C_(v,m)-R)ln2`
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

When 1 mole of an ideal monatomic gas is compressed adiabatically the intenal energy change involved is 24 cals. The temperature rise is

A gas for which lambda=1.5 is suddenly compressed to 1/4 th of the initial volume. Then the ratio of the final to initial pressure is

When a gas is compressed to half of its initial volume. If it is comparision between isothermal and adiabatic process, then a) Work done is less if the change is Isothermal b) Final pressure is more if the change is adiabatic c) Final pressure is less if the change is Isothermal

gamma for a gas is 5/3. An ideal gas at 27^@C is compressed adiabatically to 8/27 of its original volume. The rise in temperature of the gas is