Page 62 - Physical chemistry understanding our chemical world

PROPERTIES OF GASES AND THE GAS LAWS 29

pressure is p 2 and the new volume is V 2 . Equation (1.14) then holds provided the number
of moles does not vary.

Worked Example 1.4 Nitrogen gas is housed in a sealed, hollow cylinder at a pressure
3
5
of 10 Pa. Its temperature is 300 K and its volume is 30 dm . The volume within the
3
cylinder is increased to 45 dm , and the temperature is increased at the same time to
310 K. What is the new pressure, p 2 ?
We first rearrange Equation (1.14) to make the unknown volume p 2 the subject, writing

p 1 V 1 T 2
p 2 = Note how the units of
T 1 V 2
volume cancel, mean-
We insert values into the rearranged equation: ing we can employ any
unit of volume provided
3
5
10 Pa × 30 dm × 310 K the units of V 1 and V 2
p 2 = 3 are the same.
300 K × 45 dm
5
so p 2 = 0.69 × 10 Pa. The answer demonstrates how the pressure drops by about a third
on expansion.
5
SAQ 1.10 The pressure of some oxygen gas is doubled from 1.2 × 10 Pa
5
to 2.4 × 10 Pa. At the same time, the volume of the gas is decreased
3
from 34 dm 3 to 29 dm . What is the new temperature T 2 if the initial
temperature T 1 was 298 K?

Justification Box 1.1

We start with n moles of gas at a temperature T 1 , housed in a volume V 1 at a pressure
of p 1 . Without changing the amount of material, we change the volume and temperature
to V 2 and T 2 respectively, therefore causing the pressure to change to p 2 .
The number of moles remains unaltered, so we rearrange Equation (1.13) to make n
the subject:
n = p 1 V 1 ÷ RT 1
Similarly, the same number of moles n under the second set of conditions is

n = p 2 V 2 ÷ RT 2

Again, although we changed the physical conditions, the number of moles n remains
constant, so these two equations must be the same. We say

p 1 V 1 p 2 V 2
= n =
RT 1 RT 2