# Calculation of pressure & volume both isothermally & adiabatically

• es4
In summary, we discussed the isothermal and adiabatic processes in thermodynamics. We applied the Boyle equality law and adiabatic equality law to solve for the final volume in both processes. In the first scenario, using a value of 1.38 for Y, we found the final volume to be approximately 0.0262 m3. In the second scenario, with a value of 1.5 for Y, we found the final volume to be approximately 0.0262 m3. We also discussed the use of logarithms in solving for the final volume, with the preferred method being to take the nth root of y.
es4
Homework Statement
0.8 m3 of an experimental gas, compressed from a pressure of, 250 kn/m^2 to pressure of 7000 kn/m^2 . For gas in question, Cp = 1010 & Cv = 732.

Calculate the final volume of each case when gas is compressed.
>>> isothermally

Calculate the final volume when gas is adiabatically compressed & ϒ= 1.5

p1 = 250 or 250000 n/m2
p2 = 7000 or 7000000 n/m2

Cp = 1010
Cv = 732
Relevant Equations
p1v1 = p2v2 [Boyle equality law (isothermal process)]

P1 V1^ Y= P2 V2 ^Y [adiabatic Equality law]
Hi, here is my attempt... Feel free to correct me if I am doing it wrong.
v1 = .8 m3
p1 = 250 or 250000 n/m2
p2 = 7000 or 7000000 n/m2

Cp = 1010
Cv = 7321.
isothermally
p1v1 = p2v2 [Boyle equality law (isothermal process)]
250 000 * .8 = 7000000 * v2
7000000 * v2 = 250 000 * .8
v2 = 250 000 * .8/ 7000 000
v2 = 0285 m3 <<<ans

Y = Cp / Cv
Y = 1010 / 732
y = 1.38

P1 V1^Y= P2 V2^Y [adiabatic Equality law]
250 000 * .8^ 1.38 = 7000000 * v2 ^1.38
7000000 * v2 ^1.38 = 250 000 * .735
7000000 * v2 ^1.38 = 183740.2
v2 ^1.38 = 183740.2/ 7000000
v2 ^1.38 = .0262 <<< what now?

y = 1.5 [new value for gamma]

P1 V1^Y= P2 V2^Y [adiabatic Equality law]
250 000 * .8^ 1.5 = 7000000 * v2 ^1.5
7000000 * v2 ^1.5 = 250 000 * .716
7000000 * v2 ^1.38 = 178885.43
v2 ^1.38 = 178885.43/ 7000000
v2 ^1.38 = .0262 <<< what now?

If ##x^n=y##, what is x equal to? We learned how to do this in Intermediate Algebra in high school.

es4
Chestermiller said:
If ##x^n=y##, what is x equal to? We learned how to do this in Intermediate Algebra in high school.
I guess you are trying to trigger log out of me. That means, other things are correct though... :)

es4 said:
I guess you are trying to trigger log out of me. That means, other things are correct though... :)
I wouldn't use log. $$x=y^{1/n}$$

Delta2
Chestermiller said:
I wouldn't use log. $$x=y^{1/n}$$
hmm, seems like I got to brush up some old memories. Appreciated the response though.

es4 said:
hmm, seems like I got to brush up some old memories. Appreciated the response though.
Yes, I agree. In doing thermodynamics, the math is supposed to be a "gimme"

es4
Isn't ##~x^n = y~\Rightarrow~x = y^{1/n}~##taking the nth root?
Why wouldn't a log do, or an ln?

kumusta said:
Isn't ##~x^n = y~\Rightarrow~x = y^{1/n}~##taking the nth root?
Why wouldn't a log do, or an ln?
It’s just easier to calculate the value of 1/n and then raise y to that power.

I think you're right there. Just two steps away from the answer.
Using ln, just doing two computations won't give you the answer yet. You'll need to do at least three; first the value of ln##~y~##, then the value of (ln##~y##)/##n~##, then that of ##exp[{\frac {\ln y}{n}}]~## before you finally get ##x~.##

Delta2 and Chestermiller

## 1. How do you calculate pressure and volume isothermally?

The calculation of pressure and volume isothermally involves using the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and inversely proportional to its volume. This can be represented by the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

## 2. What is the formula for calculating pressure and volume adiabatically?

The formula for calculating pressure and volume adiabatically is given by the adiabatic equation, which is P1V1^γ = P2V2^γ, where γ is the adiabatic index or ratio of specific heats. This equation applies to a gas undergoing a reversible adiabatic process, meaning there is no heat exchange with the surroundings.

## 3. How does temperature affect the calculation of pressure and volume?

Temperature plays a crucial role in the calculation of pressure and volume, as it is directly proportional to pressure in an isothermal process and indirectly proportional to volume in an adiabatic process. Changes in temperature can result in significant changes in pressure and volume, which is why it is important to consider temperature in these calculations.

## 4. What are the units for pressure and volume in these calculations?

The units for pressure and volume in these calculations will depend on the units used for the gas constant, R. If R is given in SI units (J/mol*K), then pressure will be in Pascals (Pa) and volume will be in cubic meters (m^3). If R is given in other units, such as atm*L/mol*K, then pressure will be in atmospheres (atm) and volume will be in liters (L).

## 5. Can these calculations be applied to all types of gases?

Yes, these calculations can be applied to all types of gases as long as the ideal gas law or adiabatic equation can be used to model their behavior. However, it is important to note that these calculations may not be accurate for real gases at high pressures or low temperatures, as they deviate from ideal gas behavior under these conditions.

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