# Thermodynamics, ideal gas

1. Apr 2, 2012

### fluidistic

1. The problem statement, all variables and given/known data
A container contains 1g of $O_2$ at a pressure of 1 atm and temperature $T_i=47°C$. At a certain time, due to an escape of gas, the pressure is 5/8 atm and the temperature decreases to 27°C.
1)What is the volume of the container?
2)How much mass of $O_2$ has escaped?

2. Relevant equations
PV=NRT.

3. The attempt at a solution
1g of $O_2$ is worth 1/32 mol of $O_2$. Thus $V=\frac{N_iRT_i}{P_i}$. I took care of converting degrees Celcius to kelvin, took 1atm =10⁵ pascal. This gave me $V=8.31m^3$.
While $N_f = \frac{P_fV}{RT_f} \approx 208.25 mol$ which makes absolutely no sense. I should get less than 1/32, yet I get a number greater than 200. I don't know what I did wrong, I tried another approach leading to another inconsistency: $N_f=1.5 N_i$.

2. Apr 2, 2012

### I like Serena

Let's see...

$$V_i={N_i R T_i \over P_i}={\frac 1 {32} \cdot 8.31 \cdot 320 \over 10^5}=8.31 \cdot 10^{-4} m^3$$
How did you get $V_i=8.31 m^3$?

3. Apr 2, 2012

### fluidistic

Hi! Thanks Misuse of calculator apparently. I must be extremely careful to put all parenthesis I think. I now get your result.
Number of moles in the end: 0.005 or about 1/200. Thus about 0.16 g of $O_2$.

4. Apr 2, 2012

### I like Serena

Yeah, I thought as much.
And this is a problem with numbers that are "nice", so you can easily do it without calculator! ;)