# Ideal gas question with 3 variables (scary)

• jrklx250s
In summary, a horizontal, insulated cylinder contains a frictionless non conducting piston. Heat is slowly applied to the gas on the left side until the piston has compressed the gas on the right to 7.59 atm. Work is done on the gas on the right side. The final temperature of the gas on the right side is 7.59 atm and the final temperature of the gas on the left side is 273K.

## Homework Statement

A horizontal, insulated cylinder contains a frictionless non conducting piston. On each side of the piston is 54L of and inert monatomic ideal gas at 1 atm and 273K. Heat is slowly applied to the gas on the left side until the piston has compressed the gas on the right to 7.59 atm.

a) How much work is done on the gas on the right side?
b) What is the final temperature of the gas on the right side?
c) What is the final temperature of the gas on the left side?
d) How much heat was added to the gas on the left side?

## The Attempt at a Solution

My efforts...

a) Since this is a monatomic gas we can conclude that the ratio fo heat capacity is close to 5/3 so gamma = 5/3

Assuming that there is no heat transferred from left to right I can use the equation PV^Y = constant where Y = gamma

therefore I have PiVi^Y = PfVf^Y
Since I am given Pi = 1 atm and Vi = 54L and Pf = 7.59 I can rearrange for Vf...

Then I can subsititute all these values into a previous work equation i derived

W = (PfVf - PiVi)/(Y-1)
and with all known values solve for Work

OK so is this solution possible I believe I have gone wrong somewhere here

On to b)

Here I can just use the ideal gas formula PV=nRT

Assuming there is 1 mole of gas in each side I can just take the pressure and volume of the gas that I found in a) and use n= 1 and R= 0.0821

However I feel this is wrong as well...

On to c)

Here if I know the pressure and volume of the left side and the pressure volume and temperature of the right side I can just make use of the equation

PiVi/Ti = PfVf/Tf

Here I am not sure if this is right as well...

On to d)

Well since we need to calculate the heat added I must make use of the equation

Q = ΔU -W

Here we can subsitute the work in for the value found in a) if it is correct because this will be the work supplied to the right side from the left side...however I am lost in calculated ΔU is it just the change in temperatures times the heat capacity at constant pressure? Once again I am confused in calculating the ΔU because on each side the temperature volume and pressure are all changing...

Thank you very much for any help in this complicated problem.
From a very frustrated student :D.

jrklx250s said:

## The Attempt at a Solution

My efforts...

a) Since this is a monatomic gas we can conclude that the ratio fo heat capacity is close to 5/3 so gamma = 5/3

Assuming that there is no heat transferred from left to right I can use the equation PV^Y = constant where Y = gamma

therefore I have PiVi^Y = PfVf^Y
Since I am given Pi = 1 atm and Vi = 54L and Pf = 7.59 I can rearrange for Vf...
?
Then I can subsititute all these values into a previous work equation i derived

W = (PfVf - PiVi)/(Y-1)
and with all known values solve for Work
?

Your formulae are not shown. Have you written them at all?

ehild

No I didnt show my formulas because I didn't think I was even on the right track for this question. I mean I know you must use PV=nRT, PiVi^γ=PfVf^γ, and P1V1/T1 = P2V2/T2 and for the last one Q = ΔU - W...im just unsure if these are appropriate for each question

Those are the formulae you need. Go ahead.

ehild

Can you please explain how we calculate part d with the information we got from previous parts.I think for du = C_v*dT , we don't have a value for C_v.Thanks

Part A looks OK. In part B, you can't just assume that there is 1 mole. From the initial conditions, you can calculate the number of moles on each side of the piston before the heating takes place. For part C, you know the final pressure on the left side, the final volume of the left side, and the number of moles on the left side. This is enough to calculate the final temperature of the left side. To do part D, you know that the work done on the total system is zero. Therefore, the heat added has to equal the sum of the changes in internal energy for the two compartments. You know the initial and final temperatures of the two compartments, and the number of moles in each compartment, so you can calculate the changes in internal energy.

chet

## 1. What are the three variables in the ideal gas equation?

The three variables in the ideal gas equation are pressure (P), volume (V), and temperature (T). These variables are related by the equation PV = nRT, where n represents the number of moles of gas and R is the universal gas constant.

## 2. How do you calculate the pressure of an ideal gas?

The pressure of an ideal gas can be calculated using the ideal gas equation, where P = (nRT)/V. This equation shows that pressure is directly proportional to the number of moles of gas and the temperature, and inversely proportional to the volume.

## 3. Can the ideal gas equation be applied to all gases?

The ideal gas equation is based on certain assumptions, such as that the gas particles have no volume and do not exert any attractive or repulsive forces on each other. Therefore, it can only be applied to gases that behave ideally under specific conditions, such as low pressure and high temperature.

## 4. How does temperature affect the behavior of an ideal gas?

According to the ideal gas equation, temperature and pressure are directly proportional. This means that as the temperature of an ideal gas increases, its pressure will also increase. Additionally, higher temperatures can cause gas particles to move faster and collide more frequently, resulting in a higher pressure.

## 5. What is an ideal gas constant?

The ideal gas constant (R) is a proportionality constant that appears in the ideal gas equation. Its value depends on the units used for pressure, volume, and temperature. It is commonly expressed in units of Joules per mole Kelvin (J/mol·K) or atmospheres per liter Kelvin (atm/L·K).