Calculating Entropy Change in an expansion

In summary, the conversation discusses calculating the entropy change that occurs when a gas is expanded reversibly under isothermal conditions. The given information includes the initial volume, pressure, and temperature of the gas, as well as the final pressure. The formula dS= nRln (V2/V1) is used, where n represents the number of moles of gas. The final volume is calculated using the ideal gas law, and the resulting entropy change is found to be 0.34 J/K.
  • #1
iraya92
3
0
1. Suppose that you have a sample of a gas in a cylinder equipped with a piston that has a volume of 1.50 L, a pressure of 1.20 atm, and a temperature of 250 K. Suppose that the gas is expanded reversibility under isothermal conditions until the pressure is 0.75 atm. What is the entropy change that accompanies this expansion?

Homework Equations


dS= nRln (V2/V1) [/B]

The Attempt at a Solution


So I calculated V2 using p1V1=p2V2 and got 2.4L. I plugged that into the equation above but got 3.9...The answer is suppossed to be 0.34. Can you guys help me? :) [/B]
 
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  • #2
iraya92 said:
1. Suppose that you have a sample of a gas in a cylinder equipped with a piston that has a volume of 1.50 L, a pressure of 1.20 atm, and a temperature of 250 K. Suppose that the gas is expanded reversibility under isothermal conditions until the pressure is 0.75 atm. What is the entropy change that accompanies this expansion?

Homework Equations


dS= nRln (V2/V1) [/B]

The Attempt at a Solution


So I calculated V2 using p1V1=p2V2 and got 2.4L. I plugged that into the equation above but got 3.9...The answer is suppossed to be 0.34. Can you guys help me? :) [/B]
What are the units of your answer, and what are the units of the accepted answer? What value did you get for n?
 
  • #3
The units are 0.34 J/K and as for the moles I think we were to assume the gas to be one mole.
 
  • #4
Oh Okay I got it... so I was to calculate n too...duh...Thanks! Got it... *Feeling enlightened*
 
  • #5


Your calculations look correct so far. However, you may have made a mistake in plugging in the numbers into the equation. Make sure to use the natural log (ln) function on your calculator, rather than the base 10 log (log) function. Also, make sure to convert all units to their SI counterparts (i.e. atm to Pa, L to m^3) before plugging them into the equation. Doing so should give you the correct answer of 0.34 J/K.
 

Related to Calculating Entropy Change in an expansion

What is entropy change in an expansion?

Entropy change is a measure of the amount of disorder or randomness in a system that occurs during an expansion process. It is a thermodynamic property that quantifies the change in the distribution of energy within a system.

How do you calculate entropy change in an expansion?

The formula for calculating entropy change in an expansion is ∆S = Q/T, where ∆S is the change in entropy, Q is the amount of heat transferred, and T is the temperature at which the expansion occurs. This formula is valid for reversible expansion processes.

What is the unit of measurement for entropy change?

The unit of measurement for entropy change is joules per kelvin (J/K) in the International System of Units (SI). This unit is also equivalent to kilogram-meter squared per second squared per kelvin (kg·m^2·s^-2·K^-1).

Can entropy change be negative?

Yes, entropy change can be negative. This occurs when there is a decrease in disorder or randomness in a system. For example, when a gas is compressed, its particles become more ordered and the entropy change is negative.

What factors affect entropy change in an expansion?

The factors that affect entropy change in an expansion include temperature, pressure, and the type of expansion (reversible or irreversible). Higher temperatures and lower pressures generally result in higher entropy changes, while irreversible expansions may have a greater impact on entropy change due to the loss of usable energy.

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