Question

n = 4.67 mol of Hydrogen gas is initially at T = 345.0 K temperature and p_i = 3.02×10⁵ Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches p_f = 6.72×10⁵ Pa. What is the volume of the gas at the end of the compression process?

How much work did the external force perform?

How much heat did the gas emit?

How much entropy did the gas emit?

What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure?

Answer 1

(a) To solve this part, use ideal gas equation,

pf ∙ Vf = n ∙R∙Tf

=>

Vf = n∙ R∙Tf / pf

= 4.67 mol ∙ 8.3145 Pa∙m³∙K⁻¹∙mol⁻¹ ∙ 345 K / 6.72×10⁵ Pa

= 1.993 ×10⁻² m³

= 19.93 L (Answer)

(b) Reversible work done on the gas is given by the integral

....... Vf

W = - ∫ p dV

....... Vi

from ideal gas law follows for an isothermal process

p∙V = n∙R∙T = constant

=>

p = n∙R∙T/V

Hence,

................. Vf

W = -n∙R∙T∙ ∫ (1/V) dV = - n∙R∙T∙(ln(Vf) - ln(Vi)) = n∙R∙T∙ln(Vi/Vf)

................. Vi

The volume ratio can be converted to a pressure ratio:

p∙V = constant

=>

pi∙Vi = pf∙Vf

=>

Vi/Vf = pf/pi

Therefore

W = n∙R∙T∙ln(pf/pi)

= 4.67 mol ∙ 8.3145 Pa∙m³∙K⁻¹∙mol⁻¹ ∙ 345 K ∙ ln( 6.72×10⁵ Pa/ / 3.02×10⁵ Pa)

= 10.714 x 10^3 J

= 10.71 kJ (Answer)

(c) The change in internal energy of the gas equals the hat added to it plus the work done to it:

∆U = Q + W

The internal energy for an ideal gas is given by:

U = n∙Cv∙T

So if the temperature is constant, the change in internal energy is zero.

Hence,

Q + W = 0

=>

Q = - W = - 10.71 kJ

Therefore, heat emitted by the gas = 10.71 KJ (Answer)

(d) Change in entropy due a transfer of heat Q at constant thermodynamic temperature T is given by:

∆S = Q/T

For the gas in this process

∆S = -10.71×10³ J / 345 K = - 31.0 J∙K⁻¹

Therefore, entropy emitted by the gas = 31.0 J∙K⁻¹ (Answer)

(e) For an ideal gas undergoing a reversible and adiabatic process:

p∙V^γ = constant

with γ = Cp/Cv

The heat capacity ration for a diatomic ideal gas like hydrogen (H₂) is:

γ = 7/5

Relation above can rewritten in terms of pressure and temperature using ideal gas law:

V = n∙ R∙T/p

=>

p∙(n∙ R∙T/p)^γ = C

<=>

p^(1-γ) ∙ T^γ = C /(n∙ R)^γ = constant

<=>

T / p^[(γ - 1)/γ)] = [C /(n∙ R)^γ]^[1/γ] = constant

for γ = 7/5

T / p^(2/7) = constant

<=>

T' / pi^(2/7) = T / pf^(2/7)

=>

T' = T ∙ (pi/pf)^(2/7)   

= 345 K ∙ (3.02×10⁵ Pa / 6.72×10⁵ Pa)^(2/7)

= 274 K (Answer)