Question

n = 2.66 mol of Hydrogen gas is initially at T = 318 K temperature and pi = 2.49×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 9.49×105 Pa.

a.)What is the volume of the gas at the end of the compression process?

b.) How much work did the external force perform?

c.) How much heat did the gas emit?

d.) How much entropy did the gas emit?

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

Answer 1

(a) What is the volume of gas at the end of compression process?

using an ideal gas law, we have

P_f V_f = n R T_i

V_f = (2.66 mol) (8.314 J/mol.K) (318 K) / (9.49 x 10⁵ Pa)

V_f = 0.00741 m³

convert m³ to L :

V_f = 7.41 L

(b) How much work did the external force perform?

we know that, W = - P . dV                              { eq.1 }

from an ideal gas law, we have

P V = n R T   P = n R T / V

W = - (n R T) (1/V) dV

W = - (n R T) ln [(V_f - V_i)]

W = n R T ln (V_i / V_f)                                                        { eq.2 }

using a boyle's law, we have

P_i V_i = P_f V_f

V_i / V_f = P_f / P_i

THEN, we get

W = n R T ln (P_f / P_i) = (2.66 mol) (8.314 J/mol.K) (318 K) ln [(9.49 x 10⁵ Pa) / (2.49 x 10⁵ Pa)]

W = (7032.6 J) (1.33796)

W = 9409.3 J

(c) How much heat did the gas emit?

we know that, U = Q + W

The internal energy for an ideal gas is given by -

U = n C_v T

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

THEN, we get

Q = - W

Q = - 9409.3 J

That means 9.41 kJ of heat are emitted by the gas.

(d) How much entropy did the gas emit?

using a formula, we have

S = Q / T = - (9409.3 J) / (318 K)

S = - 29.5 J/K

That means 29.5 J/K are emitted.