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harina [27]
3 years ago
12

A ballon is inflated with 2.42L of helium at a temperature of 27.0°C. When put in the freezer, the volume changes to 2.37L and -

8.8°C, and the pressure is measured to be 754torr. What was the initial pressure?
Chemistry
1 answer:
natita [175]3 years ago
5 0

Answer:

838 torr  

Step-by-step explanation:

To solve this problem, we can use the <em>Combined Gas Laws</em>:

p₁V₁/T₁ = p₂V₂/T₂            Multiply each side by T₁

   p₁V₁ = p₂V₂ × T₁/T₂     Divide each side by V₁

      p₁ = p₂ × V₂/V₁ × T₁/T₂

<em>Data: </em>

p₁ = ?;             V₁ = 2.42 L; T₁ =  27.0 °C

p₂ = 754 torr; V₂ = 2.37 L; T₂ =  -8.8 °C

Calculations:

(a) Convert <em>temperatures to kelvins </em>

T₁ = (27.0 + 273.15) K = 300.15 K

T₂ = (-8.8 + 273.15) K = 264.35 K

(b) Calculate the<em> pressure </em>

p₁ = 754 torr × (2.37 L/2.42) × (300.15/264.35)  

p₁ = 754 torr × 0.979 × 1.135

p₁ = 838 torr

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A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure
Alex

Answer:

B.3/5p

Explanation:

For this question, we have to remember <u>"Dalton's Law of Partial Pressures"</u>. This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a <em>proportional relationship between moles and pressure</em>. In other words, more moles indicate more pressure and vice-versa.

P_i=P_t_o_t_a_l*X_i

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6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2

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The molar mass of oxygen gas (O_2) is 32 g/mol, so:

64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2

Now, total moles are:

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With this value, we can write the partial pressure expression for each gas:

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So, the answer would be <u>3/5P</u>.

I hope it helps!

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