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

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:

where ;

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 = -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:

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.

they vary. it used to be meat, grain, bread, candy, beverages, donuts, and fruits/vegetables. now they added exercise in.