Three moles of an ideal gas are compressed from 5.5*10^-2 to 2.5*10^-2 m^3. During the compression, 6.1*10^3J of work is done on the gas, and heat is removed to keep the temperature of the gas constant at all times. Find the temperature of the gas.

I posted this question few hours ago, and someone gave me such hints to do it. However, I don't quite get it, can anyone please explain more about it?THANKS A LOT!

p V = n R T is state equation before and after
Now work done at constant temp

dW = -p dV
but p = (n R T)/V and here n R T is constant given
so
dW = -(nRT) dV/V
so
work done = (nRT) ln(V1/V2)

by the way, that is also the heat out since internal energy depends only on T which is constant.

Stella,

I assumed that you have had calculus.
If you have not, then your book must say something like:

For an ideal gas:
With compression at constant temperature,
work in = (n R T) ln (V1/V2)
or
With compression at constant temperature
heat out = (n R T) ln (V1/V2)

Since change in internal energy = heat in - work out
and internal energy depends only on temperature
your work in is the same as the heat out
You are given the work in, so you have :

work in
n, number of moles
R gas constant
V1 and V2
so you can calculate T

work in = (n R T) ln (V1/V2)

what is the "1n" standing for?

ln is natural log. Log to the base e. It is on your calculator, probably on the same key as e^x

This must be in your physics book. Look in the index for isothermal compression or expansion or simply compression or expansion. It may be in a paragraph on "work done during volume changes".

To find the temperature of the gas, we can use the ideal gas law equation, which relates pressure, volume, number of moles, gas constant, and temperature. The equation is given as:

pV = nRT

Where p is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

In this problem, we are given that the gas is an ideal gas and the temperature is constant throughout the compression process. Therefore, the equation can be simplified to:

p1V1 = nRT

And

p2V2 = nRT

Now, let's move on to the work done on the gas. Work is defined as the transfer of energy due to the application of force over a distance. In this case, the gas is being compressed, and work is done on the gas.

The work done on the gas can be calculated using the formula:

work done = (nRT) ln(V1/V2)

Where dW is the work done, nRT is the constant given, and ln(V1/V2) is the natural logarithm of the ratio of the initial volume to the final volume.

In the problem statement, we are given that 6.1*10^3 J of work is done on the gas. So, the equation becomes:

6.1*10^3 J = (nRT) ln(V1/V2)

Since the temperature T is constant, we can solve for T by rearranging the equation:

T = (6.1*10^3 J) / (nR ln(V1/V2))

Now, substitute the given values into the equation. We are given that the initial volume (V1) is 5.5*10^-2 m^3 and the final volume (V2) is 2.5*10^-2 m^3. We are also told that there are three moles of the gas. The gas constant (R) is a universal constant and its value is 8.314 J/(mol·K).

Plug in these values into the equation to calculate the temperature (T):

T = (6.1*10^3 J) / (3 * 8.314 J/(mol·K) * ln(5.5*10^-2 m^3 / 2.5*10^-2 m^3))

By calculating this expression, you will find the temperature of the gas.