Question

Nitrogen gas in an expandable container is cooled from 45.0 C to 12.3 C with the pressure held constant at 2.58 ∗ 105 Pa. The total heat liberated by the gas is 2.58 ∗ 104 J. Assume that the gas may be treated as ideal. Find (a) the number of moles of gas; (b) the change in internal energy of the gas; (c) the work done by the gas. (d) How much heat woul

Answer 1

Answer:

(a) $n=27.14mol$

(b) $\Delta U=18.4kJ$

(c) $W=44.2kJ$

(d) $Q=-18.4kJ$

Explanation:

Hello,

(a) in this case, given the total heat, we compute the moles by using nitrogen's pressure-constant heat capacity (29.07 J/mol °C) as:

$Q=nCp\Delta T\\\\n=\frac{Q}{Cp\Delta T} =\frac{-2.58x10^4J}{29.07\frac{J}{mol\ \° C}(12.3-45.0)\° C} \\\\n=27.14mol$

(b) Next, by using nitrogen's volume-constant heat capacity (20.76 J/mol °C) we can compute its change in internal energy:

$\Delta U=nCv\Delta T=27.14mol*20.76\frac{J}{mol\ \°C}*(12.3-45.0) \°C*\frac{1kJ}{1000J} \\\\\Delta U=18.4kJ$

(c) Next, by using the firs law of thermodynamics, we compute the work done by the gas:

$W=Q-\Delta E=25.8kJ-(-18.4kJ)\\\\W=44.2kJ$

(d) Now, for isochoric processes (constant volume), since to change in volume is given, the work is null, for that reason, the heat will equal the change in internal energy of the gas, by using the first law of thermodynamics:

$Q=\Delta E+W=-18.4kJ+0\\\\Q=-18.4kJ$

And since it is negative, it also accounts for liberated heat.

Regards.

Answer 2

Answer:

The number of moles of the gas is: -27.14 mole

the charge in internal energy of the gas is -1.84 × 10⁴ J.

The work done by the gas is 4.42 × 10⁴ J

The heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.

Explanation:

The expression for the number of moles of a gas at constant pressure is as follows:

$\mathbf{n = \frac{Q}{Cp \Delta T}}$

$\mathbf{n = \frac{Q}{Cp (T_2-T_1)}}$

where ;

$C_p$ is the specific heat at constant pressure of Nitrogen gas which is = 29.07 J/mol/K

Since heat is liberated from the gas ; then:

$n = \dfrac{-2.58*10^4 }{29.07(45-12.3)}$

n = -27.14 mole

The number of moles of the gas is: -27.14 mole

b) The expression to be used in order to determine the change internal energy is:

$dU = nCv \Delta T$

where ;

n= 27.14 mole

Cv = specific heat at constant volume of Nitrogen gas = 20.76 J/mol/K

ΔT = (12.3-45)

So;

dU = (27.14)(20.76)(12.3-45)

dU = 563.426(-32.7)

dU = -18424.04328

dU = -1.84 × 10⁴ J

Thus; the charge in internal energy of the gas is -1.84 × 10⁴ J.

c)  The workdone by the gas can be calculated as;

W = Q - ΔU

W = 2.58 × 10⁴ J - (-1.84 × 10⁴ J )

W = 2.58  × 10⁴ J + 1.84  × 10⁴ J

W = 4.42 × 10⁴ J

The work done by the gas is 4.42 × 10⁴ J

d) The expression to calculated the work done is given as:

W = pdV

since the volume is given as constant ; then dV = 0

so;

W = p(0)

W = 0

Replacing 0 for W in the equation W = Q - Δ U

0 = Q - ΔU

-Q = - ΔU

Q = ΔU

Thus , the heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.