Question 5
An ideal gas initially
has pressure 1.0 × 105 Pa,
volume 4.0 × 10-4 m3 and temperature 300 K, as illustrated in Fig.
2.1.
Fig. 2.1
A change in energy of
the gas of 240 J results in an increase of pressure to a final value of 5.0 ×
105 Pa at constant volume.
The thermodynamic
temperature becomes T.
(a)
Calculate
(i)
the temperature T, [2]
(ii)
the amount of gas. [2]
(b)
The increase in internal energy ΔU
of a system may be represented by the expression
ΔU = q
+ w.
(i)
State what is meant by the symbol
1.
+q,
2.
+w.
[2]
(ii)
State, for the gas in (a), the value of
1.
ΔU,
2.
+q,
3.
+w.
[3]
[Total: 9]
Reference: Past Exam Paper – November 2016 Paper 41 & 43 Q2
Solution:
(a)
(i)
{pV
= nRT
The
change occurs at constant temperature.
The
amount of gas (number of moles) is also constant. R is the gas constant.
Taking
into account the constant quantities,
p ∝ T
We
can use proportion instead of finding each unknown quantity.}
p ∝ T or pV / T = constant or pV = nRT
{
p ∝ T
Initially,
1.0×105 ∝ 300 ------ (1)
After the change, 5.0×105 ∝ T ------ (2)
Divide
(2) by (1),
T/300 = (5.0×105)/(1.0×105)
T/300 = 5}
T (= 5 × 300 =) 1500 K
(ii)
pV = nRT
{We can use either the set of initial values of
the quantities or the set of final values.}
{Set of initial values:}
1.0×105 × 4.0×10-4 = n × 8.31 × 300
or
{Set
of final values:}
5.0×105 × 4.0×10-4 = n × 8.31 × 1500
n = 0.016 mol
(b)
(i)
1. +q: heating/thermal energy
supplied
2. +w: work done on/to system
(ii)
1. ΔU = 240 J {as given
in the question}
2. +q = 240 J {same value as given in 1. (= 240 J) and zero given for 3.}
{ΔU = q + w
+240 = q + 0
+q = +240 J}
3. +w = zero
{Since there
is no change in volume,
+w = pΔV = 0}
