A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 1400 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.0 m3 to 19.9 m3. a) Calculate the work done by the gas.
I found this to be 7.4 * 10^5 J
b) Calculate the change in internal energy of the gas
J

(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