In: Chemistry
2.20 moles of an ideal gas with CV,m=3R/2 undergoes the transformations described in the following list from an initial state described by T = 310. K and P = 1.00 bar.
The gas is heated to 685 K at a constant external pressure of 1.00 bar. Calculate q for this process.
The gas is heated to 685 K at a constant external pressure of 1.00 bar. Calculate w for this process
The gas is heated to 685 K at a constant external pressure of 1.00 bar. Calculate ΔU for this process
The gas is heated to 685 K at a constant external pressure of 1.00 bar. Calculate ΔH for this process
The gas is heated to 685 K at a constant external pressure of 1.00 bar. Calculate ΔS for this process
The gas is heated to 685 K at a constant volume corresponding to the initial volume. Calculate q for this process
The gas is heated to 685 K at a constant volume corresponding to the initial volume. Calculate w for this process
The gas is heated to 685 K at a constant volume corresponding to the initial volume. Calculate ΔU for this process.
The gas is heated to 685 K at a constant volume corresponding to the initial volume. Calculate ΔH for this process.
The gas is heated to 685 K at a constant volume corresponding to the initial volume. Calculate ΔS for this process.
The gas undergoes a reversible isothermal expansion at 310. K until the pressure is one third of its initial value. Calculate q for this process.
The gas undergoes a reversible isothermal expansion at 310. K until the pressure is one third of its initial value. Calculate w for this process.
The gas undergoes a reversible isothermal expansion at 310. K until the pressure is one third of its initial value. Calculate ΔU for this process
The gas undergoes a reversible isothermal expansion at 310. K until the pressure is one third of its initial value. Calculate ΔH for this process
The gas undergoes a reversible isothermal expansion at 310. K until the pressure is one third of its initial value. Calculate ΔS for this process
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We are given moles n = 2.2 moles
constant volume heat capacity Cv = 3R/2
Temperature T = 310 K
P = 1 bar = 0.987 atm
so, volume V = nRT/P (By ideal gas equation PV = nRT, R is gas constant , R = 0.0821 L atm/mole-K)
so, V = 2.2 moles * 0.0821 L atm /mole-K * 310 K / 0.987 atm
V = 56.73 L
So, inital volume is 56.73 L
If we increase the temperature to 685 K at constant pressure of 1 bar the volume will change as volume will be directly proportional to T if moles n and pressure P is constant.
so, Vi/Vf = Ti/Tf where subscripts i and f denote the initial and final values of volume and temperature.
so, 56.73 L/Vf = 310 K / 685 K
so, Vf = 56.73 * 685/310 L = 125.355 L
So, final volume is 125.355 L
b)
workdone w for this transformation can be calculated as
w = -PextV where Pext is the external pressure P = 0.987 atm, V = change in volume
= Vf - Vi = 125.355 L - 56.73 L = 68.625 L
so, w = -0.987 atm * 68.625 L = -67.733 L atm
As 1 L atm = 101.33 J
w = -67.733 * 101.33 J = -6863.37 J
c)
Change in internal energy U = nCvT
T = 685 K- 310 K = 375 K
n = 2.2 moles
Cv = 3R/2, R = 8.314 J/mole-K
so, U = 2.2 moles * (3/2)*8.314 J/mole-K * 375 K
= 10288.58 J
a)
By first law of thermodynamics we have
U = q + w
Where q is the heat exchanged during the process.
Putting values of U and w we get -
10288.58 J = q -6863.37 J
q = 17151.95 J
d)
Change in enthalpy of the transformation H is given as -
H = U + PV
Also, PV = -w = 6863.37 J
and U = 12969.84 J
so, H = 10288.58 J + 6863.37 J
= 17151.95 J