A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 254 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.0 m3 to 16.2 m3 . Part A Calculate the work done by the gas. Express your answer with the appropriate units. W = nothing nothing Request Answer Part B Calculate the change in internal energy of the gas. Express your answer with the appropriate units. ΔU = nothing nothing

Explanation:

Atmospheric pressure P = 10⁵ N / m ²

change in volume Δ V = 16 - 12 = 4 m³

work done by gas against constant pressure

W = P  Δ V

Putting the values above

work done = 10⁵ x 4

W = 4 x 10⁵ J .

B ) Heat added Q = 254 x 10³ cals

= 254 x 4.2 x 10³ J

= 10.66 x 10⁵ J

According to first law of thermodynamics

Q = Δ E + W

Q is heat energy added , Δ E is increase in internal energy and W is work done by gas.

Putting the values in the equation above

10.66 x 10⁵ =  Δ E +4 x 10⁵

Δ E = 6.66 x 10⁵ J .

i think its c if i'm wrong

total distance/total time

(85 + 82)/(5+3)

167/8

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