Solutions to p 21

Solutions to p 21

1. P₁V₁/(n₁T₁) = P₂V₂/(n₂T₂), but volume is constant so

P₁/(n₁T₁) = P₂/(n₂T₂),

302/(1.0*-21+273) = 320/(0.98T₂)

T₂ = 272.469 K = -0.53 ^oC

2. mass of gas = 32.4 - 32.0 = 0.4 g

PV = nRT

221.3(0.100 L) = n(8.31)(20.0 + 273)

n = .00909 moles

molar mass = m/n = 0.4 / .00909 moles = 44.0 g/mole; so yes it is possible that CO₂ was the gas, but keep in mind that it is not the only compound in the universe with such a molar mass. For example C₃H₈ gas is also 44g/mole.

3. The volume of a metal cylinder will remain constant. n will change.

P₂V = n₂RT₂

n₂ = P₂V/RT₂ = 402(3400)/(8.31*[18+273])

=565.2 moles

P₁V = n₁RT₁

n₁ = PV₁/RT₁= 452(3400)/( 8.31*[23+273])

= 624.7764009 moles

Dn = 624.78- 565.2 = 59.58 moles = 59.58 moles *32g/mole =1906 g lost or 1.91 kg

4. 1.00g/(17 g/mole) = 0.0588 moles of NH₃

ratio of water to NH₃ is 6/4, so 0.0588 moles *6/4 = 0.088 moles of water

PV = nRT

V = nRT/P = 0.088(8.31)(100+273)/ 180 = 1.51 L

5. n = PV/RT = 100*20/(8.31*(10.0 +273)) = 0.850 moles of hydrogen.

Since the ratio of hydrogen to water is 1: 1, then 0.850 moles of water will form = 15.3 g.

6. 78% = 0.78 (100) = 78 L

n = PV/RT = 99(78)/(8.31*(20.3+273)) = 3.17 moles = 3.17*28 = 88.7 g

7. PV = nRT

n/V = P/RT = 200/(8.31*(28+273) = 0.080 moles/L

0.080 moles/L *(1L / 1000ml) (38 g/mole) = 0.0030 g/ml

8. 71g of Cl₂= 1.0 mole

But if every molecule gets dissociated into two atoms, you will have twice as many moles of atoms:

Cl₂--> 2Cl

P = nRT/V = 2.0 (8.31)(25+273)/3.0 = 1651 kPa or 1.7 X 10³ kPa

9. P₁V₁/(n₁T₁) = P₂V₂/(n₂T₂)

V₂ = 0.20V₁

T₂ = 5/6T₁

n₂ = 0.90n₁

P₁V₁/(n₁T₁) = P₂(0.20V₁)/ (0.90n₁)( 5/6T₁)

P₁ = 0.20P₂ / [(5/6)*0.90]

P₂ = 3.75 P₁

Pressure has increased by a factor of 3.75