PROBLEM 66P

Use the ideal gas law to complete the table.

P |
V |
n |
T |

2.39 atm |
1.21 L |
______ |
205 K |

512 torr |
______ |
0.741 mol |
298 K |

0.433 atm |
0.192 L |
0.0131 mol |
______ |

______ |
20.2 mL |
5.71 × 10−3 mol |
20.4 °C |

Solution: Here, we are going to complete the given table.

Step1:

Boyle’s law, Charles’s law and Avogadro's law can be combined together to form a single equation which is known as ideal gas equation.

At constant T and n; V ∝ 1 / p Boyle’s law

At constant p and n; V ∝ T Charles’ law

At constant p and T; V ∝ n Avogadro’s law

Thus,

V ∝ nT / p

V = R

Where R is the proportionality constant called Universal gas constant. On rearranging the above equation we get,

pV = nRT ----(1)

Equation (1) is called ideal gas equation.

Step2:

The value of R is 8.20578 x 10-2 L atm K-1 mol-1

Using this value, let us complete the given table.

P |
V |
n |
T |

2.39 atm |
1.21 L |
0.1719 mol |
205 K |

512 torr |
26.88 L |
0.741 mol |
298 K |

0.433 atm |
0.192 L |
0.0131 mol |
77.3 K |

6.877 atm |
20.2 mL |
5.71 × 10−3 mol |
20.4 °C |

- Given, p = 2.39 atm

V = 1.21 L

T = 205 K

From equation (1), n = pV / RT

= (2.39 atm x 1.21 L) / (8.20578 x 10-2 L atm K-1 mol-1 x 205 K)

= (2.8919 atm L) / (16.82 L atm mol-1)

= 0.1719 mol.

b) Given, p = 512 torr = 512 x 0.00131579 atm = 0.674 atm

n = 0.741 mol

T = 298 K

From equation (1), V = nRT / p

= (0.741 mol x 8.20578 x 10-2 L atm K-1 mol-1 x 298 K) / (0.674 atm)

= (18.12 L atm) / (0.674 atm)

= 26.88 L.

c) Given, p = 0.433 atm

V = 0.192 L

n = 0.0131 mol

From equation (1), T = pV / nR

= (0.433 atm x 0.192 L) / (0.0131 mol x 8.20578 x 10-2 L atm K-1 mol-1)

= (0.083136 atm L) / (0.001075 L atm K-1)

= 77.3 K(approx.).

d) Given, V = 20.2 mL = 20.2 / 1000 L= 0.020 L

n = 5.71 x 10-3 mol

T = 20.4 oC = 20.4 + 273.15 K = 293.55 K

From equation (1), p = nRT / V

= (5.71 x 10-3 mol x 8.20578 x 10-2 L atm K-1 mol-1 x 293.55 K) / (0.020 L)

= (0.13754 L atm) / (0.020 L)

= 6.877 atm.

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