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
