Question

If the final pressure on the gas is 1.20 atm , calculate the entropy change for the process. Check lecture notes or textbook for the formula connecting the change of entropy and gas volume. Assume that neon behaves as an ideal gas in this experiment.

Answer 1

Answer: $-2.34\frac{J}{K}$

Explanation:

In thermodynamics, entropy (symbolized as S) is a physical quantity for a thermodynamic system in equilibrium. It measures the number of micro states compatible with the macro state of equilibrium, it can also be said that it measures the degree of organization of the system, or that it is the reason for an increase in internal energy versus an increase in the temperature of the system.

If we assume we have 0.6 mol of an ideal gas at 350 K at an initial pressure of 0.75 atm, we calculate the entropy change as:

$S=nRln\frac{P1}{P2}$

Where S is entropy, n is the number of moles, R is the gas constant R is 8.314 J / mol·K, P1 is the initial pressure and P2 is the final pressure. Then we substitute the values and solve for S.

$S= (0.6 mol).(8.3145\frac{J}{mol.k})(ln\frac{0.75atm}{1.2atm})$

$S= -2.34\frac{J}{K}$

Answer 2

The question is not complete and the complete question is ;

the pressure on 0.600 mol of an ideal gas at 350 K is increased isothermally from an initial pressure of 0.750 atm (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process.

Answer:

Entropy change = -2.34 J/k

Explanation:

In thermodynamics, the entropy of a system is expected to increase with increase in temperature, increase in volume, or increase in number of gas particles.

Now, the gas is expanded isothermally and so its temperature is constant. Thus, as pressure of an ideal gas increases, the number of microstates possible for a system decreases and this results in the decrease of reaction entropy and thus is negative.

Thus, ΔS = nRIn(P1/P2)

Now, from the question,

n = 0.06 moles

P1 = 0.75 atm and P1=1.2 atm

R is gas constant and is 8.3145 J/kg.mol

Thus, ΔS = 0.06 x 8.3145 x In(0.75/1.2) = -2.34 J/k