Question 7
(a) The volume of an ideal gas in a cylinder is 1.80 × 10-3 m3 at a pressure of 2.60 × 105 Pa and a temperature of 297 K, as
illustrated in Fig. 2.1.
Fig. 2.1
The thermal energy required to raise the temperature by
1.00 K of 1.00 mol of the gas at
constant volume is 12.5 J.
The gas is heated at constant volume such that the
internal energy of the gas increases
by 95.0 J.
(i) Calculate
1. the amount of gas, in mol, in the cylinder, [2]
2. the rise in temperature of the gas. [2]
(ii) Use your answer in (i) part 2 to
show that the final pressure of the gas in the
cylinder is 2.95 × 105 Pa. [1]
(b) The gas is now allowed to expand. No thermal energy
enters or leaves the gas.
The gas does 120 J of work when expanding against the
external pressure.
State and explain whether the final temperature of the
gas is above or below 297 K. [3]
Reference: Past Exam Paper – June 2013 Paper 42 Q2
Solution:
(a)
(i)
1.
pV = nRT
2.60×105 × 1.80×10-3 = n × 8.31 × 297
Number
of moles, n = 0.19 mol
2.
{The thermal energy
required to raise the temperature of 1 kg of a substance by 1K is called
the ‘specific heat capacity’.
We use the formula: Heat Δq = mcΔT where
m is the mass
But here, instead of mass,
we have the number of moles and similarly, we have the corresponding amount of
thermal energy for 1 mole (instead of mass). So, the formula can be modified to
be:
Δq = ncΔT
where n is the number of
moles and c is the heat required to raise the temperature of 1 mole of the gas
by 1 K.}
Δq = ncΔT
95.0 = 0.190 × 12.5 × ΔT
ΔT = 40 K
(ii)
{Since the volume is
constant, (pressure law:)}
p / T = constant
{p1 / T1 = p2 / T2}
{Final temperature =
Initial temperature + ΔT}
(2.6×105) / 297 = p / (297+40)
Pressure p = 2.95×105 Pa
(b)
{First law of
thermodynamics: ΔU = ΔQ + ΔW
ΔQ is
the amount of heat/energy (positive
when heat is given TO the system and negative when heat is taken FROM
system)
ΔU is
the change in internal energy (positive when internal energy of system increases and negative
when internal energy of system decreases.
Note: ΔU = KE + PE
For an ideal gas, PE = 0.
So, ΔU = KE
The kinetic energy depends
on temperature (KE = 3/2 kT). A change in internal energy of the ideal gas is shown
by a change in temperature – this is the energy of the gas itself
ΔW is
the work done (positive
when external work is done ON system and negative when external work is done BY
system.
When a gas expands, it
does work against external forces. ΔW = -ve
When a gas contracts, work
is done on the gas. ΔW = +ve
No thermal energy enters
or leaves the gas: ΔQ = 0
The gas does 120J of work
when expanding against the external pressure: ΔW = -120J
Replacing the energy
values,
ΔU = ΔQ + ΔW = 0 – ΔW = -
120 J}
ΔU = – ΔW = –120 J
As ΔU is negative, the internal energy decreases.
Since the internal energy represents the KE of the molecules of the ideal gas, the
KE of the molecules decreases. Thus, the temperature lowers.
Thank you. Can you please explain 21/m/j/2017 question 2 next? (Related to forces and equilibrium).
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