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Ronch [10]
3 years ago
7

PLEASE PLEASE HELP ME!! I WILL BRAINLIST YOU!! A 28.0 g sample of N2 is in a rigid 4.50 L container at 32 °C. Calculate the pres

sure in the flask in torr.
Chemistry
1 answer:
chubhunter [2.5K]3 years ago
6 0

Answer:

P=4184.36 torr

Explanation:

For this problem we can use the idea gas law,

PV=nRT

where P is pressure, V is volume, n is moles of substance, R is the constant, and T is temperature, We will need to manipulate it and a couple values so that we can get our answer in torr. To do this, let's rearrange the equation to solve for P.

P=\frac{nRT}{V}

Next, let's convert T (32) to kelvin. To do this, add 273 to the value.

32+273=305

Next, let's convert 28g of N2 to moles of N2. To do this, divide 28 by the molar mass of N2 (28.02)

\frac{28.0}{28.01} \\=.99 moles of N2

Next, we need to find out what value of R we need to use. Because you want your answer in torr, we will need to use the value of

R=62.36\frac{torr}{mol*K}

Now that we have all of our values, let's plug them into the equation (I'll be excluding units for simplicity but they cancel out to leave units of torr in the answer)

P=\frac{(.99)(62.36)(305)}{4.50}\\P=4184.36

P=4184.36 torr

<em>Very briefly</em>, I don't know what your periodic table looks like, but mine has nitrogen's molar mass as 14.01. If you have a different mass of nitrogen, this pretty drastically impacts the value. For example, if your nitrogen's molar mass was rounded to 14.0 (making N2 28g) then it would be one mole instead of .99 moles, raising the answer to 4228.70 torr. If you have any different values just plug them in in their proper spots. A similar concept applies to the Kelvin conversion.

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Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

P\cdot V = n\cdot R\cdot T,

where

  • P is the pressure of the gas,
  • V is the volume of the gas,
  • n is the number of gas particles in the gas,
  • R is the ideal gas constant, and
  • T is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume V of the gas. Rearrange the ideal gas equation for volume:

V = \dfrac{n \cdot R \cdot T}{P}.

Both the temperature of the gas, T, and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

  • T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K},
  • T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}.

The volume of the gas is proportional to its temperature if both n and P stay constant.

  • n won't change unless the balloon leaks, and
  • consider P to be constant, for calculations that include T.

V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}.

Now, keep the temperature at T_1 =258.65\;\text{K} and change the pressure on the gas:

  • P_1 = 0.995\;\text{atm},
  • P_2 = 0.720\;\text{atm}.

The volume of the gas is proportional to the reciprocal of its absolute temperature \dfrac{1}{T} if both n and T stays constant. In other words,

V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}

(3 sig. fig. as in the question.).

See if you get the same result if you hold T constant, change P, and then move on to change T.

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