140 moles of a gas which has y = 1.31 expands from Vi = 0.5m³ to = 1.8m³ following the relation P(V) = 2.0 × 10³ Pa m³ / +1 +1.2 x 105. Pa. m³ V (2)
140 moles of a gas which has y = 1.31 expands from Vi = 0.5m³ to = 1.8m³ following the relation P(V) = 2.0 × 10³ Pa m³ / +1 +1.2 x 105. Pa. m³ V (2)
Chapter3: The First Law Of Thermodynamics
Section: Chapter Questions
Problem 62P: For a temperature increase of 10 at constant volume, what is the heat absorbed by (a) 3.0 mol of a...
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![140 moles of a gas which has y = 1.31 expands from V₂ = 0.5m³ to
Vf = 1.8m³ following the relation
P(V) = 2.0 × 105 Pa
Pa
m³-
m³1/1/ +1.2 x 105:
m³
30. What is Cv, the molar specific heat at constant volume, for this gas?
(a) 1.3R
(b) 2.3R
(c) 3.2R
(d) 4.2R
(e) 5.2R](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37589965-6822-4c51-9fa5-c1e77c2c8460%2F9e1a03f9-170d-47bd-ab6d-e41ee6246d1e%2Frkpxk0h_processed.png&w=3840&q=75)
Transcribed Image Text:140 moles of a gas which has y = 1.31 expands from V₂ = 0.5m³ to
Vf = 1.8m³ following the relation
P(V) = 2.0 × 105 Pa
Pa
m³-
m³1/1/ +1.2 x 105:
m³
30. What is Cv, the molar specific heat at constant volume, for this gas?
(a) 1.3R
(b) 2.3R
(c) 3.2R
(d) 4.2R
(e) 5.2R
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