Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Assuming that water vapour is an ideal gas, the internal energy change (\Delta U) when 1 mol of water is vapourised at 1 bar pressure and 1000C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol-1 and R = 8.3 J mol-1 K-1) will be

  • Option 1)

    41.00 kJ mol-1

  • Option 2)

    4.100 kJ mol-1

  • Option 3)

    3.7904 kJ mol-1

  • Option 4)

    37.904 kJ mol-1

 

Answers (1)

best_answer

As we learnt in

Molar heat capacity for isobaric process C(p) -

dH=n\: C_{P}\: dT
 

- wherein

dH=dE+d(PV)

or

dH=dE+n\, R\, dT

 

 Given \Delta H=41KJmol^{-1}=41000KJmol^{-1}

T=100^{\circ}C=273+100=373K

for n = 1

\Delta U=\Delta H-\Delta nRT=41000-(2\times8.314\times373)

=\:37898.88\:Jmol^{-1}\:=37.9\:KJmol^{-1}


Option 1)

41.00 kJ mol-1

This option is incorrect.

Option 2)

4.100 kJ mol-1

This option is incorrect.

Option 3)

3.7904 kJ mol-1

This option is incorrect.

Option 4)

37.904 kJ mol-1

This option is correct.

Posted by

prateek

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 chemistry needs conceptual understanding; 5 tips to score high
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon; paper pattern, exams from January 22
anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters

Latest Question