Answer:
a) the final volume is 38.285 L
b)
w = - 1839.158 J,
q = 0,
ΔU = - 1839.158 J
Explanation:
Given that;
number of moles n = 2
initial pressure p1 = 3 atm
external pressure p_ext = 1 atm
temperature T1 = 350 k
we know that gas constant R = 8.315 JK⁻¹mol¹ = 0.08206 L atm K⁻¹ mol⁻¹
a)
to determine the initial volume , we use the equation for ideal gas;
V1 = nRT1/p1
so we substitute
V1 = (2 × 0.08206 × 350) / 3
V1 = 57.442 . 3
V1 = 19.147 L
given that, the gas is expanded irreversibly and adiabatically until the volume has doubled;
so Final volume v2 will be;
V2 = 2 × V1
V2 = 2 × 19.147 L
V2 = 38.285 L
Therefore, the final volume is 38.285 L
b)
to determine w, q and U;
we know that, work done in gas expansion at a constant speed is expressed in the following equation;
w = -p_ext ΔV
where ΔV is change in volume
so we substitute
w = -( 1 atm) ( 38.285 L - 19.147 L)
w = -( 1 atm) ( 19.138 L)
w = -(19.138 L-atm)
we know that 1 L-atm = 101.325 J.
so
w = -(19.138 × 101.325 J)
w = - 1839.158 J
now, since there is no transfer of heat into or out of the system for the adiabatic process;
q = 0
using the first law of thermodynamics, changes in internal energy will be;
ΔU = q + w
we substitute
ΔU = 0 + ( - 1839.158 J )
ΔU = - 1839.158 J
