# What pressure is required to reduce 75 mL of a gas at standard conditions to 11 mL at a temperature of 27◦C? Answer in units of atm.?

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- az_lenderLv 72 months ago
If "standard conditions" means 1 atm and 0C, the answer will be

(1 atm)*(75/11)*(303.15/273.15)

= about 7 atm but use a calculator.

If "standard conditions means 1 bar and 0C, the answer will be

(1 bar)(75/11)(303.15/273.15)*(1 atm/1.013 bar)

= again near 7 atm.

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- Roger the MoleLv 72 months ago
(1 atm) x (27 + 273) K / (273 K) x (75 mL / 22 mL) = 3.7 atm

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- Jim MoorLv 72 months ago
PV = nRT <<<memorize, you'll need it.

Since nR are constant, PV/T init = PV/T final

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