Find the volume of 100 g of CO2 at 28 C and 11.50 torr ........ i need a solution ....plzzzzzzzzzzzzzzzz help me.........
Answers
Answer:
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Answer:
Similarly, the temperature change alone will increase the volume, since volume and
temperature are directly proportional (CHARLES' LAW). Therefore, the correction fraction for
temperature must make the volume larger and is 298 K / 273 K.
The solved problem looks like this:
P T 760 mm Hg 298 K
V2 = V1 x x = 125 mL x x = 230 mL
P T 450 mm Hg 273 K
correction fractions
In many problems only one variable changes; these problems are much easier to solve.
EXAMPLE: A gas sample has a pressure of 742 mm Hg at 25o
C. Calculate the pressure of the
gas at 100o
C.
Instead of blindly substituting in equation (4) let us reason. Equation (4) is given below without
numerical subscripts on the V and T fractions: they are simply considered to be correction
fractions: the second fraction will be the first multiplied by two correction fractions, one for
volume and one for temperature. No mention is made of the gas volume so we assume it is
constant and the volume correction fraction is not needed.
The temperature changes from 298 K to 373 K; this temperature increase will increase the
pressure (GAY-LUSSAC'S LAW). Therefore the correction fraction for temperature must make
the pressure larger and is 373 K / 298 K. The solved problem looks like this:
V T T 373 K
P2 = P1 x x = P1 x = 742 mm Hg x = 929 mm Hg
V T T 298 K
MASS-VOLUME RELATIONSHIPS: PVT problems work for any gas. If we want to know the
volume of a given mass of gas, or the mass of a given volume of gas, then we must consider
a specific gas; the mass of 100 mL of oxygen is different from the mass of 100 mL of neon.
We must first evaluate R, the molar gas constant. Solve the IDEAL GAS LAW for R and
substitute known values for n, P, V, and T: one mole of a gas occupies 22.4 L at 1 atm and
273 K:
PV (1 atm)(22.4 L) L-atm
R = = = 0.0821
nT (1 mol)(273 K) mol-K