Asked by HELP PLS

Given R=0.081L*atm/mol*K

What is the volume of a mole of gas at a pressure of 3.12 atm and a temperature of 270 K? The answer is 7.1g. Show your work of how you solves this problem.
Answered by Anonymous
P V = nR T
P = 3.2 atm = 324,240 Pascals
T = 270 K
n = 1
R = 8.314
so
V = n R T / P = 1 * 8.314 * 270 / 324,240 = 0.00692 meters^3 = 6.92 Liters
Answered by DrBob222
7.1 g is a volume???I don't think so.
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