Hello, Brooklyn,
The quantity of gas does not change (no gas added or removed), so we can use a simplified verion of the ideal gas law:
P1V1/T1 = P2V2/T2
It's a long story, but this is a combination of the ideal gas law for states 1 and 2, where both n (the moles) and R (the gas constant) both cancel. It is a handy equation to memorize for cases where the moles of gas do not change.
Next, rearrange the equation to solve for the unknown, V2, in this case:
V2 = V1(T2/T1)(P1/P2)
Standard tempearature and pressure are 1.0 atm and 273.15K (or 0C). These become the values for P2 and T2. Convert the temperatures into degrees K (add 275.15) to the temperature value, and make a table with clearly marked values, and their untis. Then enter those in the equation.
Please note that I organized the expression for V2 to illustrate that the new volume will be the initial volume times the ratios of of initial and final pressures and temperatures. One is an inverse ratio (P1/P2). Plug in the numbers, cancel the units, and if the only unit left is volume (in this case, ml) then we can start feeling good that we've taken the correct steps. But does the result make sense? We took a fixed amount of gas and changed its pressure and temperature. The volume is allowed to change, so think of this gas in a balloon. The tempearture is lowered, so that should shrink the balloon. The pressure is raised by quite a bit. Our equation says that the volume is dependent on the pressure ratio (P1/P2). That makes sense - the volume should decrease as the pressure in increased. Since this is a ratio (0.38/1), I can see, with no calculator, that this alone will reduce the volume by more than half. So we should expect an answer with a volume less than half it's original.
The value I get is 14.9ml, 15 ml with 2 sig figs. That is a little less than half of the starting pressure, so I'm happy, and on to other things.
Bob
