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nadya68 [22]
3 years ago
11

A sample of gas has a pressure of 3.00 atm at 25 degrees Celsius. What would the pressure be at 52 degrees Celsius if the volume

stays constant? Which gas law does this problem represent?
Chemistry
1 answer:
Doss [256]3 years ago
8 0
The law that relate pressure and temperature of gases at constant volumes is Gay-Lussac's Law.

It states that the pressure of a fixed mass of gas is directly proportional to the absolute temperature when the volume is constant.

 P / T = constant => P1 / T1 = P2 / T2

=> P2 = T2 * P1 / T1

Remember that the formula uses absolute temperatures.

T2 = 52 + 273.15 = 325.15 K

T1 = 25 + 273.15 = 298.15 K

=> P2 = 325.15K * 3.00 atm / 298.15K = 3.27 atm.

Answer: 3.27 atm
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What mass of sucrose (C12H22O11) should be combined with 546 g of water to make a solution with an osmotic pressure of 8.80 atm
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To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

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\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 8.80 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (sucrose) = ?

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 564 mL    (Density of water = 1 g/mL)

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

T = Temperature of the solution = 290 K

Putting values in above equation, we get:

8.80atm=1\times \frac{\text{Mass of sucrose}\times 1000}{342.3\times 546}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 290K\\\\\text{Mass of sucrose}=\frac{8.80\times 342.3\times 546}{1\times 1000\times 0.0821\times 290}=69.08g

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The equation for this reaction is

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