A sample consisting of 65.0 g of xenon is confined in a container at 2.00 atm and 298 K and then allowed to expand adiabatically (a) reversibly to 1.00 atm, (b) against a constant pressure of 1.00 atm. Calculate the final temperature and the expansion work at each case. Use the fact that xenon is a monoatomic gas.

[a]. - 445.45J,

[b]. - 367.92 J

Explanation:

The following parameters are given in the question above. These information are used in solving this problem.

The mass of Xenon = 65.0 g of xenon, pressure = 2.00 atm, temperature = 298 K.

The number of moles of xenon = mass/ molar mass = 65g/ 131.293= 0.495.

The cp= 3/2 R, cp =3/2R + R = 5/2 R.

j = cp/cv = 3/2.

[a]. The final temperature,T2 = (2)^-2/3 × (298)^5/3 = T2^5/3.

Final temperature,T2 = 225.84K.

Expansion work = nCv [ T₂ - T₁] = 0.495 × 3/2 × 8.314 × [ 225.84 - 298] = - 445.45J.

(b). The final temperature can be Determine as;

3/2( T2 - 298k) = - 1 (T2 /1 - 298/2).

3/2(T2 - 298) = - T2 + 149K.

3T2 - 894 = - 2T2 +298K.

T2 = 238.4 K.

Workdone= nCv (T2 - T1) = 0.495 × 3/2 × 8.314 (238.4 - 298) = - 367.92 J.

Answer:

Step-by-step explanation:

The input force is 50 N. But it will not create not any change. No mechanical advantage is observed.

Answer:

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the periodic table)

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 ×328 K

Solving:

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.