Question

REVERSIBLE EXPANSION OF 0.5 mol OF AN IDEAL GAS Wmax IS -3.918KJ.THE VALUE OF ΔU IS

Answer 1

Solution:-

Work done in an isothermal reversible process is given by-

W=−nRTln  

V  

1

​  

 

V  

2

​  

 

​  

 

Where,

n = no. of moles =1

R= molar gas constant =8.314Jmol  

−1

K  

−1

 

T= temperature =300K

V  

1

​  

=10dm  

3

 

V  

2

​  

=20dm  

3

 

∴W=−1×8.314×300×ln  

10

20

​  

 

⇒W=−2494.2×(ln2−ln1)=−1728.8 J=−1.73 kJ

Now, from first law of thermodynamics,

ΔU=ΔH+W

Since temperature is constant, i.e. ΔU=0.

∴ΔH=−W=−(−1.73)=+1.73KJ