Q#7 (Past Exam Paper – June 2013 Paper 42 Q2)
(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]
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.
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