**Solutions to p 21**

1. P_{1}V_{1}/(n_{1}T_{1})
= P_{2}V_{2}/(n_{2}T_{2}), but volume is
constant so

P_{1}/(n_{1}T_{1})
= P_{2}/(n_{2}T_{2}),

302/(1.0*-21+273)
= 320/(0.98T_{2})

T_{2} = 272.469 K = -0.53 ^{o}C

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 CO_{2} was the gas, but keep in mind that it is not the only compound in the universe
with such a molar mass. For example C_{3}H_{8} gas is also
44g/mole.

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

P_{2}V =
n_{2}RT_{2}

n_{2} =
P_{2}V/RT_{2} = 402(3400)/(8.31*[18+273])

=565.2 moles

P_{1}V =
n_{1}RT_{1}

n_{1} =
PV_{1}/RT_{1}= 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 NH_{3}

ratio
of water to NH_{3} 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 Cl_{2 }= 1.0 mole

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

Cl_{2}--> 2Cl

P
= nRT/V = 2.0 (8.31)(25+273)/3.0 = 1651 kPa or 1.7 X 10^{3} kPa

9. P_{1}V_{1}/(n_{1}T_{1})
= P_{2}V_{2}/(n_{2}T_{2})

V_{2} = 0.20V_{1}

T_{2} = 5/6T_{1}

n_{2} = 0.90n_{1}

P_{1}V_{1}/(n_{1}T_{1})
= P_{2}(0.20V_{1})/ (0.90n_{1})( 5/6T_{1})

P_{1} = 0.20P_{2} / [(5/6)*0.90]

P_{2} = 3.75 P_{1}

Pressure has increased by a factor of 3.75