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ANTONII [103]
3 years ago
12

A student combines a sample of gas (2.0 L) at 3.5 atm with with another gas (1.5 L) at 2.8 atm pressure into an empty 7.0 L flas

k. Assuming the gases are combined at constant temperature, what is the total gas pressure (in atmospheres) in the 7.0 L flask?
Chemistry
1 answer:
Sedbober [7]3 years ago
4 0

Answer:

Total gas pressure is 1.60 atm

Explanation:

To solve this question we can use the Ideal Gases Law. We need to determine how many moles of each gas will be finally present at the flask of 7 L.

Let's asume the gas, are at Asbsolute T°, 273K

P. V = n . R . T

3.5 atm . 2L = n . 0.082 . 273K

(3.5 atm . 2L) / (0.082 . 273K) = 0.313 moles

(2.8 atm . 1.5L) / (0.082 . 273K) =  0.188 moles

Total moles = 0.313 mol + 0.188 mol = 0.501 mol

Let's calcualte the hole pressure

P . 7L = 0.501 moles . 0.082 . 273K

P = (0.501 moles . 0.082 . 273K) / 7L → 1.60 atm

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Read 2 more answers
A 1.93-mol sample of xenon gas is maintained in a 0.805-L container at 306 K. Calculate the pressure of the gas using both the i
Alex

Answer : The pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

Explanation :

First we have to calculate the pressure of gas by using ideal gas equation.

PV=nRT

where,

P = Pressure of Xe gas = ?

V = Volume of Xe gas = 0.805 L

n = number of moles Xe = 1.93 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of Xe gas = 306 K

Now put all the given values in above equation, we get:

P\times 0.805L=1.93mole\times (0.0821L.atm/mol.K)\times 306K

P=60.2atm

Now we have to calculate the pressure of gas by using van der Waals equation.

(P+\frac{an^2}{V^2})(V-nb)=nRT

P = Pressure of Xe gas = ?

V = Volume of Xe gas = 0.805 L

n = number of moles Xe = 1.93 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of Xe gas = 306 K

a = pressure constant = 4.19L^2atm/mol^2

b = volume constant = 5.11\times 10^{-2}L/mol

Now put all the given values in above equation, we get:

(P+\frac{(4.19L^2atm/mol^2)\times (1.93mole)^2}{(0.805L)^2})[0.805L-(1.93mole)\times (5.11\times 10^{-2}L/mol)]=1.93mole\times (0.0821L.atm/mol.K)\times 306K

P=44.6atm

Therefore, the pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

5 0
3 years ago
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