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11111nata11111 [884]
3 years ago
14

Consider 1.5 L of air in a patents lungs at 37.0 C and 1.00 atm pressure. What is the volume would this air occupy if it were at

25 C under a pressure of 5.00 x 10^2 atm?
Chemistry
1 answer:
SSSSS [86.1K]3 years ago
5 0

Answer:

0.00288 L = 2.88 mL.

Explanation:

  • To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant.

T is the temperature of the gas in K.

  • If n is constant, and have two different values of (P, V and T):

<em>P₁V₁T₂ = P₂V₂T₁</em>

<em></em>

P₁ = 1.0 atm, V₁ = 1.5 L, T₁ = 37°C + 273 = 310 K.

P₂ = 500.0 atm, V₂ = ??? L, T₂ = 25°C + 273 = 298 K.

<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (1.0 atm)(1.5 L)(298 K)/(500.0 atm)(310 L) = <em>0.00288 L = 2.88 mL.</em>

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2. How much heat is released when 432g of water cools down from 710C to 180C?
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Acetic acid has a Kb of 2.93 °C/m and a normal boiling point of 118.1 °C. What would be the boiling point of a solution made by
alexdok [17]

Boiling point elevation is given as:

ΔTb=iKbm

Where,

ΔTb=elevation in the boiling point

that is given by expression:

ΔTb=Tb (solution) - Tb (pure solvent)

Here Tb (pure solvent)=118.1 °C

i for CaCO3= 2

Kb=2.93 °C/m

m=Molality of CaCO₃:

Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)

=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)

=(100.0÷100 g/mol)/0.4

= 2.5 m

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ΔTb=iKbm

=2×2.93×2.5

=14.65 °C

boiling point of a solution can be calculated:

ΔTb=Tb (solution) - Tb (pure solvent)

14.65=Tb (solution)-118.1

Tb (solution)=118.1+14.65

=132.75

3 0
3 years ago
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