Hello, Dawson,

The gas law equation, PV=nRT, can be used to answer these questions, but in cases where the moles of gas stays the same, a simpler exression is:

P1V1/T1 = P2V2/T2

where 1 and 2 are the initial and final states. respectively.

(The n and R both cancel out since their values are constant between the two states, __when gas is neither added nor removed__).

We'll assume a convenient volume of gas of 1 liter. We want to see what V2 will be under the new condifions.

Temperatures MUST be in Kelvin (add 273 to C)

Rearrange the equation for V2 and enter the values:

V_{2} = (P_{1}V_{1}T_{2})/(P_{2}T_{1})

V_{2} = (1.04atm*1Liter**(215+273)K)/(7.08atm*(122+273))

V_{2} =

The pressure is increased from 1.04 atm to 7.08 atm, while the temperature is decreased from 215°C to 122°C. Before we calculate, let's guestimate what the answer might look like. We're increasing pressure by almost a factor of 7, and decreasing temperature from 488K to 395K, a factor of around 20%. Both would result in a smaller volume. I'll simply guess around an 80% reduction in volume, which would mean we go from 1 liter to 200 ml.

If we get an anser significantly different, we should double check our work. My calculation comes to 0.181 liters, or 181 ml. CLOSE! So I'm happy to move on.

Bob