Question 2
2
3 marks
Question 2(b)
2(b)
The mean kinetic energy \(<E_{\mathrm{K}}>\) of a molecule of an ideal gas is given by the expression where k is the Boltzmann constant and T is the thermodynamic temperature of the gas. A cylinder contains 1.0 mol of an ideal gas. The gas is heated so that its temperature changes from 280 K to 460 K .
structured3 marks
Question 2(b)(ii)
2(b)(ii)
During the heating, the gas expands, doing \(1.5 \times 10^{3} \mathrm{~J}\) of work. State the first law of thermodynamics. Use the law and your answer in (i) to determine the total energy supplied to the gas. total energy = J
Mediumstructured3 marks
Answer
increase in internal energy = heat supplied + work done on system 2240 = energy supplied - 1500 energy supplied \(=3740 \mathrm{~J}\)
Question 2
2
the rate h of thermal energy gained by the ice from the surroundings. W
structured2 marks
Question 2(c)
2(c)
Explain why the change from C to A involves external work and a change in internal energy.
Hardstructured2 marks
Answer
temperature changes/decreases so internal energy changes/decreases volume changes (at constant pressure) so work is done
Question 2
2
the rise in temperature of the gas. temperature rise = K
structured2 marks
Question 2(b)
2(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 .
Mediumstructured3 marks
Answer
change in internal energy is \(120 \mathrm{~J} / 25 \mathrm{~J} \quad\) B1 internal energy decreases / \(\underline{\Delta} U\) is negative / kinetic energy of molecules decreases M1 so temperature lower A1
Question 2
2
A cylinder contains 5.12 mol of an ideal gas at pressure \(5.60 \times 10^{5} \mathrm{~Pa}\) and volume \(3.80 \times 10^{-2} \mathrm{~m}^{3}\).
structured4 marks
Question 2(c)
2(c)
4 marks
Question 2(c)(i)
2(c)(i)
Use the information in (b)(i) to calculate the external work done during the expansion of the gas.
Mediumstructured2 marks
Answer
\(W=p \Delta V\) C1 \[ \begin{aligned} W=5.60 \times 10^{5} \times(4.75-3.80) \times 10^{-2} W=5320 \mathrm{~J} \end{aligned} \] A1
Question 2(c)(ii)
2(c)(ii)
Use the first law of thermodynamics to determine the total thermal energy transferred to the gas in (b). Explain your reasoning. energy = J
Hardstructured2 marks
Answer
work done on the gas is negative or work is done by the gas B1 increase in internal energy = thermal energy transferred to gas + work done on gas so thermal energy transferred to gas =7980-(-5320) \(=13300 \mathrm{~J}\) A1
Question 2
2
A constant mass of an ideal gas has a volume of \(3.49 \times 10^{3} \mathrm{~cm}^{3}\) at a temperature of \(21.0^{\circ} \mathrm{C}\). When the gas is heated, 565 J of thermal energy causes it to expand to a volume of \(3.87 \times 10^{3} \mathrm{~cm}^{3}\) at \(53.0^{\circ} \mathrm{C}\). This is illustrated in Fig.2.1.
structured4 marks
Question 2(b)
2(b)
The pressure of the gas is \(4.20 \times 10^{5} \mathrm{~Pa}\). For this heating of the gas,
structured4 marks
Question 2(b)(i)
2(b)(i)
calculate the work done by the gas, work done = J
Easystructured2 marks
Answer
work done \(=p \Delta V\) (do not allow use of V instead of \(\Delta V\) )
Question 2(b)(ii)
2(b)(ii)
use the first law of thermodynamics and your answer in (i) to determine the change in internal energy of the gas. change in internal energy = J
Mediumstructured2 marks
Answer
increase/change in internal energy = heating of system + work done on system C1 =565-160 \(=405 \mathrm{~J}\)
Question 2
2
the period of Deimos in its orbit about Mars. period = hours
structured3 marks
Question 2(c)
2(c)
2 marks
Question 2(c)(ii)
2(c)(ii)
Use the equation in (b)(ii) to explain that, for an ideal gas, a change in internal energy \(\Delta U\) is given by where \(\Delta T\) is the change in temperature of the gas.
Mediumstructured2 marks
Answer
no intermolecular forces so no potential energy (change in) internal energy is (change in) kinetic energy and this is proportional to (change in ) T
Question 2
2
The product of the pressure p and the volume V of an ideal gas is given by the expression where m is the mass of one molecule of the gas.
structured4 marks
Question 2(c)
2(c)
A cylinder contains 1.0 mol of an ideal gas.
structured4 marks
Question 2(c)(i)
2(c)(i)
The volume of the cylinder is constant. Calculate the energy required to raise the temperature of the gas by 1.0 kelvin. energy = J
Mediumstructured2 marks
Answer
either energy required \(=(3 / 2) \times 1.38 \times 10^{-23} \times 1.0 \times 6.02 \times 10^{23} \quad\) C1 or
Question 2(c)(ii)
2(c)(ii)
The volume of the cylinder is now allowed to increase so that the gas remains at constant pressure when it is heated. Explain whether the energy required to raise the temperature of the gas by 1.0 kelvin is now different from your answer in (i).
Hardstructured2 marks
Answer
energy is needed to push back atmosphere/do work against atmosphere so total energy required is greater
Question 2
2
5 marks
Question 2(b)
2(b)
A fixed mass of an ideal gas at a temperature of \(20^{\circ} \mathrm{C}\) is sealed in a cylinder by a piston, as shown in Fig. 2.1. The initial volume of the gas is \(1.24 \times 10^{-4} \mathrm{~m}^{3}\). Thermal energy is supplied to the gas and its volume increases by \(5.20 \times 10^{-5} \mathrm{~m}^{3}\).
structured2 marks
Question 2(b)(i)
2(b)(i)
The piston is freely moving so that the gas is always at atmospheric pressure. Atmospheric pressure is \(1.01 \times 10^{5} \mathrm{~Pa}\). Calculate the work done by the gas. work done by gas = J
Easystructured2 marks
Answer
\(W=p \Delta V\) C1 \[ \begin{aligned} =1.01 \times 10^{5} \times 5.20 \times 10^{-5} =(+) 5.25 \mathrm{~J} \end{aligned} \] A1
Question 2(c)
2(c)
The gas in (b) is allowed to return to its starting temperature. The piston is now fixed in position. Thermal energy is supplied to increase the temperature to the same final temperature as in (b). Use the first law of thermodynamics to suggest and explain how the specific heat capacity of the gas for this situation compares with the value in (b)(iii).
Hardstructured3 marks
Answer
no change in volume so no work is done (by the gas) B1 (same temperature change so) same change in internal energy B1 less thermal energy needs to be supplied so c is less B1
Question 2
2
air and soft tissue.
structured3 marks
Question 2(b)
2(b)
A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig. 2.1.
structured3 marks
Question 2(b)(i)
2(b)(i)
State the change in internal energy of the gas during one complete cycle P Q R P. change =
Mediumstructured0 marks
Answer
zero A1
Question 2(b)(ii)
2(b)(ii)
Calculate the work done on the gas during the change from P to Q. work done =
Mediumstructured0 marks
Answer
work done \(=p \Delta V \quad\) C1 \(=4.0 \times 10^{5} \times 6 \times 10^{-4}\) \(=240 \mathrm{~J} \quad\) (ignore any sign)
Question 2(b)(iii)
2(b)(iii)
Some energy changes during the cycle PQRP are shown in Fig. 2.2. Complete Fig. 2.2 to show all of the energy changes.
Hardstructured3 marks
Answer
change work done / J heating / J increase in internal energy / J \(P \rightarrow Q\) + 2 4 0 -600 - 3 6 0 \(Q \rightarrow R\) 0 +720 + 7 2 0 \(R \rightarrow P\) - 8 4 0 +480 - 3 6 0 (correct signs essential) (each horizontal line correct, 1 mark - max 3) B3 [3]
Question 2
2
A fixed mass of an ideal gas has a volume V and a pressure p. The gas undergoes a cycle of changes, X to Y to Z to X, as shown in Fig. 2.1. Table 2.1 shows data for p, V and temperature T for the gas at points X, Y and Z .
structured5 marks
Question 2(d)
2(d)
5 marks
Question 2(d)(i)
2(d)(i)
The first law of thermodynamics for a system may be represented by the equation State, with reference to the system, what is meant by: \(\Delta U:\) q : W :
Mediumstructured3 marks
Answer
\((\Delta U:)\) increase in internal energy (of the system) B1 ( q :) thermal energy supplied to the system B1 ( W :) work done on system B1
Question 2(d)(ii)
2(d)(ii)
Explain how the first law of thermodynamics applies to the change Z to X .
Mediumstructured2 marks
Answer
volume increases and work is done by the gas B1 temperature decreases and internal energy decreases B1