Question

2.60 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.

Calculate:

A. The gas is heated to 710 K at a constant external pressure of 1.00 bar. Calculate q for this process.

B. The gas is heated to 710 K at a constant external pressure of 1.00 bar. Calculate w for this process.

C. The gas is heated to 710 K at a constant external pressure of 1.00 bar. Calculate ΔU for this process..

D. The gas is heated to 710 K at a constant external pressure of 1.00 bar. Calculate ΔH for this process.

E. The gas is heated to 710 K at a constant external pressure of 1.00 bar. Calculate ΔS for this process.

F. The gas is heated to 710 K at a constant volume corresponding to the initial volume. Calculate q for this process.

G. The gas is heated to 710 K at a constant volume corresponding to the initial volume. Calculate w for this process.

H. The gas is heated to 710 K at a constant volume corresponding to the initial volume. Calculate ΔU for this process.

I. The gas is heated to 710 K at a constant volume corresponding to the initial volume. Calculate ΔH for this process.

J. The gas is heated to 710 K at a constant volume corresponding to the initial volume. Calculate ΔS for this process.

K. 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.

L. 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.

M. 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..

N. 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.

O. 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.

Thank you

Answer 1

We are given moles n = 2.6 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 = 2.6 moles * 0.0821 L atm /mole-K * 310 K / 0.987 atm

V = 67.044 L

So, inital volume is 67.044 L

If we increase the temperature to 710 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, 67.044 L/Vf = 310 K / 710 K

so, Vf = 67.044 * 710/310 L = 153.55 L

So, final volume is 153.55 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 = 153.55 L - 67.044 L = 86.506 L

so, w = -0.987 atm * 86.506 L = -85.381 L atm

As 1 L atm = 101.33 J

w = -85.381 * 101.33 J = -8651.657 J

c)

Change in internal energy U = nCvT

T = 710 K- 310 K = 400 K

n = 2.6 moles

Cv = 3R/2, R = 8.314 J/mole-K

so, U = 2.6 moles * (3/2)*8.314 J/mole-K * 400 K

= 12969.84 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 -

12969.84 J = q -8651.657 J

q = 21621.497 J

d)

Change in enthalpy of the transformation H is given as -

H = U + PV

Also, PV = -w = 8651.657 J

and U = 12969.84 J

so, H = 12969.84 J + 8651.657 J

= 21621.497 J