Question

Suppose 2.60 mol of an ideal gas of volume V 1 = 3.50 m 3 at T 1 = 290 K is

allowed to expand isothermally to V 2 = 7.00 m 3 at T 2 = 290 K. Determine:

a) the work done by the gas, [Ans = 4,350 J]

b) the heat added to the gas, and [Ans = take a guess]

c) the change in internal energy of the gas. [Ans = check definition of iso…]

Answer 1

a) $Workdone = nRT \ln { V_1 \over V_2}$

$2.6 × 8.314 × 290( \ln{7 \over 3.5}) = 4385.62J$

b) since it is Isothermal, there is no internal energy. Then, $\Delta E_{int} = 0$

$\Delta E_{int} = Q-W$

$W= Q= 4385.62J$

c) $\Delta E_{int} = 0$ (since it is isothermic)