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A 3.00 L flask contains 2.33 g of argon gas at 312 mm Hg What is the temperature of the gas

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debooxpp

Answer:

T= 257.36 k

Explanation:

using the ideal gas law

pv=nRt

first, convert pressure from mmhg to kpa

312 x (101.3\760)= 41.58 kpa

R is constant= 8.31

to get n(number of moles)

n=m\M          (m is mass, M is molar mass)

molar mass of argon is 39.948

n= 2.33\39.948

n=0.0583

substitute;

41.58 x 3 = 0.0583 x 8.31 x T

T= (41.58 x 3)\ (0.0583 x 8.31)

T= 257.36 k

abidemiokin

The temperature of the gas under ideal conditions is 257.36K

In order to get the temperature of the gas, we will use the ideal gas equation expressed according to the formula:

[tex]PV = nRT[/tex]

P is the pressure of the gas = 312mmHg = 41.58KPa

V is the volume of the gas = 3.00L

n is the moles of the gas =  0.1165moles

R is boltzmann constant = 8.31

T is the required temperature

Mole = mass/molar mass

Mole = 2.33/40

Mole of argon = 0.05825moles

Substitute the given parameters into the formula

[tex]T=\frac{PV}{nR}\\ T=\frac{41.58 \times 3}{0.05825\times8.31}\\T=257.36K[/tex]

Hence the temperature of the gas under ideal conditions is 257.36K

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