Answer:
(a) The work done by the gas 8.005 x 10⁵ J
(b) the change in internal energy is 5.0575 x 10⁶ J
Explanation:
Given;
heat added to the gas, Q = 1400 kcal = 1400 kcal x 4184 J/kcal = 5.858 x 10⁶ J.
change in volume, ΔV = 19.9 m³ - 12.0 m³ = 7.9 m³
atmospheric pressure, P = 101325 N/m²
(a) The work done by the gas = PΔV
= 101325 x 7.9
= 8.005 x 10⁵ J
(b) the change in internal energy is obtained from first law of thermodynamic;
ΔU = Q - W
ΔU = 5.858 x 10⁶ J - 8.005 x 10⁵ J
ΔU = 58.58 x 10⁵J - 8.005 x 10⁵ J
ΔU = 5.0575 x 10⁶ J