# Homework Help: Formula help

1. Dec 30, 2004

### Jimsac

Not sure what formula to use to solve this problem
A cylinder of gas at room temperature has a pressure of p1. To what temperature in degrees C would the termperature have to be increased for the pressure to be 1.5p1?

Thank you all for your help

2. Dec 30, 2004

### arildno

Well, what law do you think might be applied here?

3. Dec 30, 2004

### Jimsac

p NT
V

4. Dec 30, 2004

### dextercioby

I'm sorry Arildno,but your question is a little bit vague...
Here's how i'd do it:
a)Do you know what an ideal gas is??
b)Is the gas in your problem ideal ?
c)If so,do you know the thermal state law for an ideal gas??
d)If so,apply it for the 2 states given in your problem,taking into account the fact that the tank/cylinder has constant volume.

Daniel.

5. Dec 30, 2004

### Gokul43201

Staff Emeritus
I don't see how this is a formula. Surely your text covers the gas laws/kinetic theory of gases.

6. Dec 31, 2004

### Ameen Khan Gauzel

Hi there. Well, if you think carefully, you might find the answer logically. Though you might use the formula PV = nRT (for the same gas), it might be more lenthy to arrive at the answer. You should know that the pressure of a gas varies directly with its temperature.

Hence, you might use the formula P = kT, where k is a constant. So, if the know the initial pressure (p1) and final pressure (p2), and on top of that, you know the initial temperature (assume c1), you might substitute in the formula above.

You will get two equations (one relating the initial pressure and temperature, and the other one relating the final temperature and pressure). If you solve these equations simultaneously, you will arrive at the answer. OK.

7. Jan 2, 2005

### Jimsac

ok so i know that my intial pressure is room temperature 20degress C It has a pressure of p1. It is increased to 1.5p1.
Do I take the work done by the system + the internal work?

8. Jan 2, 2005

### Q_Goest

The ideal gas law is an equation of state. If certain values are known, you can solve for the others. Determine what variables you know, and what variable you need to solve for.

Note also that since you don't know mass (or moles) you might solve for that, then knowing that the mass (or moles) is the same after the temperature change, solve for pressure.

Work and energy are not needed for this calculation.

9. Jan 3, 2005

### Ameen Khan Gauzel

You don't need to take into account internal energy or work done by the gas Jimsac. All you need to do is to solve these 2 equations: -

but you will need to convert the temperatures in Kelvin, using K = C + 273.15

So, room temperature becomes(initial temperature), c1 = 20 + 273.15 = 293.15 in Celcius

so,

p1 = k(293.15)...........[equation 1]

and,

1.5(p1) = k(T).............[equation 2], T is the final temperature in Kelvin

you can do a division. you can divide equation [2] by [1]

so, [2] / [1] : -

1.5(p1) / (p1) = kT / k(293.15)

p1 and k get cancelled, and you will be left with: -

1.5 = T/293.15

and T = 439.725 Kelvin

and the final temperature is (439.725 - 273.15) Kelvin = 166.575 inCelcius.

You don't need to determine mass or number of moles for this question.