Home
Class 11
CHEMISTRY
Two moles of a triatomic linear gas (neg...

Two moles of a triatomic linear gas (neglect vibration degree of freedom) are taken through a reversible process ideal starting from A as shown in figure.
The volume ratio `(V_(B))/(V_(A))=4`. If the temperature at A is `-73^(@)C`, then :

(ii) Total enthalpy change in both steps is :

A

3000 R

B

4200 R

C

2100 R

D

0

Text Solution

Verified by Experts

The correct Answer is:
B, C

(i) (c) `w=-P.DeltaV=-nRDeltaT=-2xx8.314xx600`
`=-9.97kJ`
(ii) (b) `DeltaH_("total")=DeltaH_(AB)+DeltaH_(BC)=nC_(p,m)DeltaT+0`
`=2xx(7)/(2)xxRxx(800-200)`
= 4200 R
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    NARENDRA AWASTHI|Exercise Level 3 - One Or More Answers Are Correct|1 Videos
  • THERMODYNAMICS

    NARENDRA AWASTHI|Exercise Level 3 - Match The Column|2 Videos
  • THERMODYNAMICS

    NARENDRA AWASTHI|Exercise Level 1 (Q.121 To Q.150)|1 Videos
  • STOICHIOMETRY

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

Similar Questions

Explore conceptually related problems

A sample of an ideal gas with initial pressure 'P' and volume 'V' is taken through an isothermal process during which entropy change is found to be DS. The work done by the gas is

The variation of pressure P with volume V for an ideal monoatomic gas during an adiabatic process is shown in figure. At point A the magnitude of rate of change of pressure with volume is

One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperature at A,B and C are 400K, 800K and 600K respectively. Choose the correct statement:

One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1l to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300K then total entropy change of system in the above process is: [R =0.082L atm mol^(-1)K^(-1)= 8.3J mol^(-1) K^(-1) ]