# A sample of gas occupies 1.55L at 27.0C and 1.00atm pressure whats the volume if the pressure rise to 50.0atm?

You can approximate a solution by assuming that the gas remains "ideal". Use the formula PV=nRT for this.

P=Pressure
V=Volume
n=number of moles
R=gas constant
T=Temperature

Since n, R and T are constant then the right side of the equation does not change. Therefore PV must be the same in both cases. Since P=50 times greater than it was before, V must be 50 times smaller (1.55L/50=0.031L). Notice that several shortcuts were taken: 1) no unit conversion since it can be factored out and eliminated 2) You are making a HUGE assumption by using the "ideal" gas formula. However, given the amount of information that you have provided this is the only way I can think of.
P1V1=P2*V2
1.00atm*1.55L=50.0atm*V
V=0.031L=31mL
PV = nRT

T and n are held constant. Giving you the temperature was slightly (and intentionally) misleading.

Therefore PV = a constant

Therefore P1V1 = P2V2

P1 = 1 atm V1 = 1.55 L P2 = 50 atm V2 = SOLVE
Boyle's Law. Inverse proportion at constant temp.
Pressure Increase <===> Volume Decrease.
P1 x V1 = P2 x V2
1.0atm รท 50atm = 0.02
1.55L x 0.02 = 0.031L = 31mL.

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