A gas cylinder initially contained 49.5L of Ar gas at a pressure of 85.0 atm and...
A gas cylinder initially contained 49.5L of Ar gas at a pressure of 85.0 atm and a temperature of 24oC. Then 1250g of Ar gas were removed from the cylinder. What is the new pressure of the gas, which remains in the cylinder? The cylinder volume and temperature do not change.
*** Show your work THE ANSWER IS 69.6atm but please show how to get there thanks!!
Homework Answers
V1 = initial volume = 49.5 L
P1 = 85 atm , T1 = 24C = 24+273 = 297 K
we first calculate moles of Ar
PV = nRT is the equation where R = 0.08206 liter atm/molK
85 x 49.5 = n x 0.08206 x 297
n = 172.0586
Ar gas removed = 1250 g , we convert it to moles
Moles = mass of Ar / Atomic mass of Ar = 1250 / 39.95 = 31.29
Now Ar moles left after removing 31.29068 moles = 172.0586-31.29068 = 140.77
Now they mentioned V , t are same , we find P
P x 49.5 = 140.768 x 0.08206 x 297
P = 69.3 atm ( rounding of figures in calculation reason for slightl difference)
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