Answer:
Explanation:
2.5 g of He = 2.5 / 4  mole 
= .625 moles 
9 g of oxygen = 9/32 
= .28 mole of oxygen 
C_p of He = 3/2 R
C_p of O₂ = 5/2 R
A ) Initial thermal energy of He = 3/2 n R T
= 1.5 x .625 x 8.32 x 300
= 2340 J
Initial thermal energy of O₂ = 5/2 n R T
= 2.5 x .28 x 8.32 x 620 
= 3610.88 J 
B ) If T be the equilibrium temperature after mixing 
gain of heat by helium 
= n C_p Δ T 
= .625 x 3/2 R x ( T - 300 )
Loss of heat by oxygen 
n C_p Δ T 
= .28 x 5/2 R x ( 620 - T )
Loss of heat = gain of heat 
.625 x 3/2 R x ( T - 300 ) = .28 x 5/2 R x ( 620 - T )
1.875 T- 562.5 = 868- 1.4 T 
3.275 T = 1430,5 
T = 436.8 K 
Thermal energy of He 
= 1.5 x .625 x 8.32 x 436.8
= 3407 J
 thermal energy of O₂
= 2.5 x .28  x 8.32 x 436.8
= 2543.92 J
C )
Heat energy transferred
=  .28 x 5/2 R x ( 620 - T )
=  .28 x 5/2 x  8.32 x ( 620 - 436.8 )
1066.95 J
 Heat will flow from O₂ to He
Final temperature is 436.8 K