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A cylinder, with a piston pressing down with a constant pressure, is filled with 1.90moles of a gas (n1), and its volume is 48.0L (V1). If 0.800mole of gas leak out, and the pressure and temperature remain the same, what is the final volume of the gas inside the cylinder?

AND

A sample of gas in a cylinder as in the example in Part A has an initial volume of 70.0L , and you have determined that it contains 2.00moles of gas. The next day you notice that some of the gas has leaked out. The pressure and temperature remain the same, but the volume has changed to 17.5L . How many moles of gas (n2) remain in the cylinder?

Respuesta :

IthaloAbreu

Answer:

Part A: 27.8 L; Part B: 0.500 mol

Explanation:

Part A

For an ideal gas, we can use the equation:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

In the beggining:

P*48.0 = 1.90*RT

P/RT = 1.90/48.0

P/RT = 0.03958

For constants pressure and temperature, after 0.800 mol of gas leaks out: n = 1.90 - 0.800 = 1.10 mol

PV = 1.10*R*T

P/RT = 1.10/V

0.03958 = 1.10/V

V = 1.10/0.03958

V = 27.8 L

Part B

In the beginning

PV = nRT

P*70.0 = 2.00*RT

P/RT = 2.00/70.0

P/RT = 0.02857

After the gas leaked out:

PV = nRT

P*17.5 = nRT

P/RT = n/17.5

0.02857 = n/17.5

n = 0.500 mol

Giangiglia

Answer:

A) [tex]V_2=27.8 L[/tex]

B) [tex]n_2=0.5 mol[/tex]

Explanation:

Part 1

Ideal gas equation:

[tex]P*V_1=n_1*R*T[/tex]

[tex]V_1=48 L[/tex]

[tex]n_1=1.9 mol[/tex]

P, T and R are constants.

[tex]\frac{P}{R*T}=\frac{n_1}{V_1}=\frac{1.9 mol}{48 L}=0.0396 mol/L[/tex]

Now:

[tex]n_2=1.9 mol - 0.8 mol=1.1 mol[/tex]

[tex]V_2=\frac{n_2}{P/R*T}[/tex]

[tex]V_2=\frac{1.1 mol}{0.0396 mol/L}=27.8 L[/tex]

Part 2

[tex]P*V_1=n_1*R*T[/tex]

[tex]V_1=70 L[/tex]

[tex]n_1=2 mol[/tex]

P, T and R are constants.

[tex]\frac{P}{R*T}=\frac{n_1}{V_1}=\frac{2 mol}{70 L}=0.0286 mol/L[/tex]

Now:

[tex]V_2=17.5 L[/tex]

[tex]n_2=V_2*\frac{P}{R*T}[/tex]

[tex]n_2=17.5 L*0.0286 mol/L=0.5 mol[/tex]

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