This question is in **two** parts. **Part 1** is about ideal gases and specific heat capacity. **Part 2** is about simple harmonic motion and waves.

**Part 1** Ideal gases and specific heat capacity

a State **two** assumptions of the kinetic model of an ideal gas.[2]

(i) Define what is meant by the term *one mole of argon*.

(ii) The temperature of the argon is 300 K. The piston is fixed and the argon is heated at constant volume such that its internal energy increases by 620 J. The temperature of the argon is now 350 K.

Determine the specific heat capacity of argon in J kg^{–1 }K^{–1} under the condition of constant volume. (The molecular weight of argon is 40)[4]

Explain, in terms of the molecular model of an ideal gas, why the temperature of argon decreases on expansion.[3]

**▶️Answer/Explanation**

## Markscheme

a

point molecules / negligible volume;

no forces between molecules except during contact;

motion/distribution is random;

elastic collisions / no energy lost;

obey Newton’s laws of motion;

collision in zero time;

gravity is ignored;

(i) the molecular weight of argon in grams / 6.02×10^{23} argon

atoms / same number of particles as in 12 g of C-12;*(allow atoms or molecules for particles)*

(ii) mass of gas = 0.040kg ;

specific heat = \(\frac{Q}{{m\Delta T}}\)* or* 620 = 0.04×

*c*×50;

*(i.e. correctly aligns substitution with equation)*

\( = \left( {\frac{{620}}{{0.040 \times 50}} = } \right)310{\rm{Jk}}{{\rm{g}}^{{\rm{ – 1}}}}{{\rm{K}}^{{\rm{ – 1}}}}\);

c.

temperature is a measure of the average kinetic energy of the molecules;*(must see “average kinetic” for the mark)*

energy/momentum to move piston is provided by energy/momentum of molecules that collide with it;

the (average) kinetic energy of the gas therefore decreases; *Do not allow arguments in terms of loss of speed as a result of collision with a moving piston.*

This question is about internal energy and thermal energy (heat).

a Distinguish between internal energy and thermal energy.[3]

*θ*of the kiln. The iron is then quickly transferred to water held in a thermally insulated container. The water is stirred until it reaches a steady temperature. The following data are available.

Thermal capacity of the piece of iron = 60JK^{–1}Thermal capacity of the water = 2.0×10^{3}JK^{–1}Initial temperature of the water = 16°C

Final temperature of the water = 45°C

The thermal capacity of the container and insulation is negligible.

*θ*and the above data, for the energy transfer of the iron in cooling from the temperature of the kiln to the final temperature of the water.

*θ*. [4]

**▶️Answer/Explanation**

## Markscheme

a internal energy is the total kinetic and potential energy of the molecules of a body;

thermal energy is a (net) amount of energy transferred between two bodies;

at different temperatures;

*θ*−45];

(ii) (2.0×10^{3}×29)=5.8×10^{4}J;

*θ*−45]=5.8×10

^{4};

*θ*=1000°C; (

*allow 1010°C to 3 sig fig*)

The first scientists to identify alpha particles by a direct method were Rutherford and Royds. They knew that radium-226 (\({}_{86}^{226}{\text{Ra}}\)) decays by alpha emission to form a nuclide known as radon (Rn).

a Write down the missing values in the nuclear equation for this decay.

[1]

At the start of the experiment all the air was removed from cylinder B. The alpha particles combined with electrons as they moved through the wall of cylinder A to form helium gas in cylinder

B. The wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.[1]

^{15}alpha particles to be emitted during the experiment. The experiment was carried out at a temperature of 18 °C. The volume of cylinder B was 1.3 x 10

^{–5}m

^{3}and the volume of cylinder A was negligible. Calculate the pressure of the helium gas that was collected in cylinder B.[3]

**▶️Answer/Explanation**

## Markscheme

a 222 * AND *4

*Both needed.*

**OR**

alpha particles have a low penetration power**OR**

thin glass increases probability of alpha crossing glass**OR**

decreases probability of alpha striking atom/nucleus/molecule

*p* = 4.5 x 10^{–9} x 8.31 x «\(\frac{{2.91}}{{1.3 \times {{10}^{ – 5}}}}\)»

**OR**

*p* = 2.7 x 10^{15} x 1.38 x 10^{–23} x «\(\frac{{2.91}}{{1.3 \times {{10}^{ – 5}}}}\)»

0.83 or 0.84 «Pa»

energy levels are discrete

wavelength/frequency of photon is related to energy change * or *quotes

*E*=

*hf*= \(\frac{{hc}}{\lambda }\)

**or**Eand is therefore also discrete

**OR**

means that readers have confidence in the validity of work

*OWTTE*