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Explanation:
hope that helps
Answer
given,
heat added to the gas,Q = 3300 kcal
initial volume, V₁ = 13.7 m³
final volume, V₂ = 19.7 m³
atmospheric pressure, P = 1.013 x 10⁵ Pa
a) Work done by the gas
W = P Δ V
W = 1.013 x 10⁵ x (19.7 - 13.7)
W = 6.029 x 10⁵ J
b) internal energy of the gas = ?
now,
change in internal energy
Δ U = Q - W
Q = 3300 x 10³ cal
Q = 3300 x 10³ x 4.186 J
Q = 1.38 x 10⁷ J
now,
Δ U = 1.38 x 10⁷ - 6.029 x 10⁵
Δ U = 1.32 x 10⁷ J
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