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Aleks04 [339]
3 years ago
6

Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa

gage, and the atmospheric pressure is 100 kPa absolute. Some time later, after considerably more gas is pumped into the tank, the pressure in the tank is 212 kPa gage, and the temperature is still 13°C. What will be the ratio of the mass of natural gas in the tank when p = 212 kPa gage to that when the pressure was 117 kPa gage?
For this situation in which the tank volume is the same before and after filling, which of the following is the correct relation for the ratio of the mass after filling M2 to that before filling M1 in terms of gas temperatures T1 and T2 and pressures p1 and p2?
a. M2/M1= p2T2/p1T1
b. M2/M1= p1T2/p2T1
c. M2/M1= p2T1/p1T2
d. M2/M1= p1T1/p2T2
1. What is the absolute pressure in the tank before filling?
2. What is the absolute pressure in the tank after filling?
3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?
Chemistry
1 answer:
drek231 [11]3 years ago
6 0

Answer:

1.  the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c (\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} })

Explanation:

To find  -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 117 kPa

⇒Absolute pressure ( p1 )  = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 212 kPa

⇒Absolute pressure (p2)  = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. V_{1} = V_{2}

As we know that, P ∝ M

⇒ \frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }

⇒\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }

⇒\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378 ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒\frac{P}{T} ∝ M

⇒\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }

So, The correct relation is c option.

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Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
ohaa [14]
<span>Answer: 0.094%


</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x


3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177


4) Solve the equation:


You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:


</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M


5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
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