Answer:
ΔE
,₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
Step-by-step explanation:
Given the data in the question;
T
= 300 K, T
= 412 K, n = 0.7 mol
since helium is monoatomic;
Cv = (3/2)R, Cp = (5/2)R
W₁ = 0 J [ at constant volume or ΔV = 0]
Now for the first cylinder; from the first law of thermodynamics;
Q₁ = ΔE
,₁ + W₁
Q₁ = ΔE
,₁ = n × Cv × ΔT
we substitute
Q₁ = ΔE
,₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )
Q₁ = ΔE
,₁ = 0.7 × 12.471 × 112
Q₁ = ΔE
,₁ = 977.7 J
Therefore, ΔE
,₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J