If 8.4 moles of Ne occupies 25 L of volume and exerts a pressure of 2.07 atm, what is the temperature of the gas? *
A.75.0 °C
B.75.0 K
C.318 °C
D.348 K
E.7612 K
Answers
Answered by
0
Answer:
V(ideal)=
P
RT
=
15
0.0821×250
=1.368
V(real)=
12
V(ideal)
=
12
1.368
∴z=
V(ideal)
V(real)
=
12×1.368
1.368
=0.083
Here, z (compression factor) <1,
Therefore, attractive forces are dominant.
Answered by
1
with the help of ideal gas equation PV =nRT
2.07× 25 = 8.4 × 0.0821 × T
T= 2.07 × 25 / 8.4 × 0.0821 = 75 K
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