1)

Two moles of an ideal monoatomic gas occupies a volume V at 27° C. The gas expands adiabatically to a Volume 2 V. Calculate(i) the final temperature of the gas and (ii) change in its internal energy.


A) (i)189 K (ii) 2.7 KJ

B) (i) 195 K (ii) -2.7 KJ

C) (i) 189 K (ii) -2.7 KJ

D) (i) 195 K (ii) 2.7 KJ

Answer:

Option C

Explanation:

For adiabatic process relation of temperature and volume is,

T2Vγ12=T1Vγ11

T2(2V)23=300(V)23

           [ γ=53 for monoatomic gases ]

T2=300223189K

 Also, in adiabatic process,

  Q=0,U=W

 OR       U=nR(T)γ1

  =2×32×253(300189)

    2.7KJ

     T2189K,U2.7KJ