Suppose 2.00 mol of an ideal, monatomic gas is initially at a pressure of 3.00 atm and a temperature T = 350 K. It is expanded irreversibly and adiabatically (q = 0) against a constant external pressure of 1.00 atm until the volume has doubled. A. Calculate the final volume.
B. Calculate w, q and ?U for this process , in joules.

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

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.