Chapter 5 Quizzes | Topic 5 Solids, Liquids and Gases

Quiz 1 — Units and $\rho=m/V$

1. The SI unit of density is A. $\text{g/cm}^3$ B. $\text{N/m}^2$ C. $\text{kg/m}^3$ D. $\text{kg/m}^2$

Specification 5.1 and 5.3. $\text{g/cm}^3$ is common in the lab but not the SI unit. $\text{N/m}^2$ is the pascal.

2. $75\,\text{cm}^3$ in $\text{m}^3$ is A. $7.5\times 10^{-2}\,\text{m}^3$ B. $7.5\times 10^{-5}\,\text{m}^3$ C. $7.5\times 10^{-3}\,\text{m}^3$ D. $0.75\,\text{m}^3$

Divide $\text{cm}^3$ by $10^6$. $75\div 10^6=7.5\times 10^{-5}$.

3. $2.4\,\text{g/cm}^3$ in $\text{kg/m}^3$ is A. $2400\,\text{kg/m}^3$ B. $2.4\,\text{kg/m}^3$ C. $0.0024\,\text{kg/m}^3$ D. $24\,\text{kg/m}^3$

Multiply g/cm³ by $1000$. Water $1.0\,\text{g/cm}^3=1000\,\text{kg/m}^3$.

4. Mass $0.180\,\text{kg}$, volume $7.5\times 10^{-5}\,\text{m}^3$. Density is A. $2.4\,\text{kg/m}^3$ B. $24\,\text{kg/m}^3$ C. $1.35\times 10^{-5}\,\text{kg/m}^3$ D. $2400\,\text{kg/m}^3$

$\rho=m/V=0.180/(7.5\times 10^{-5})=2400\,\text{kg/m}^3$. C inverted the fraction.

5. A cube of side $2.0\,\text{cm}$ has mass $16\,\text{g}$. Its density is A. $500\,\text{kg/m}^3$ B. $2000\,\text{kg/m}^3$ C. $2.0\,\text{kg/m}^3$ D. $8.0\,\text{kg/m}^3$

$V=8.0\,\text{cm}^3=8.0\times 10^{-6}\,\text{m}^3$, $m=0.016\,\text{kg}$, $\rho=2000\,\text{kg/m}^3$. Or $16/8=2.0\,\text{g/cm}^3=2000\,\text{kg/m}^3$.

Quiz 2 — Density practical

1. The volume of a regular rectangular block is found from A. length $\times$ width $\times$ height, measured with a ruler or callipers B. the overflow from a eureka can only C. the mass on a balance D. $p=h\rho g$

A eureka can is for irregular shapes. Mass is not volume.

2. In a eureka-can method the volume of an irregular rock is A. the volume of water left in the can B. the mass of the rock in grams C. the volume of water that overflows into the measuring cylinder D. the reading on the balance after the rock is wet

Displaced volume equals object volume (provided it sinks fully).

3. Air bubbles stuck to a submerged object make the measured density A. too high, because the overflow is too small B. too low, because the overflow is too large C. unchanged, because bubbles have no volume D. infinite

Bubbles add extra displaced volume, so $V$ is too big and $\rho=m/V$ is too small.

4. An object floats on water. To find its volume with a eureka can you should A. push it under with your finger and ignore the extra overflow B. use only the overflow while it floats C. dry it and estimate the volume by eye D. attach a sinker and subtract the sinker’s overflow from the combined overflow

A floating object only displaces its own weight of water, not its volume.

5. When reading a measuring cylinder you should A. view the bottom of the meniscus at eye level B. view the top of the meniscus from above C. shake the cylinder until the surface is flat D. estimate halfway between the highest and lowest marks

Parallax from above or below is a classic error mark on 5.4.

Quiz 3 — $p=F/A$ and pressure in fluids

1. $1\,\text{Pa}$ is the same as A. $1\,\text{N/m}$ B. $1\,\text{N/m}^2$ C. $1\,\text{kg/m}^3$ D. $1\,\text{N}$

Specification 5.5. Pressure is force per unit area.

2. $F=400\,\text{N}$, $A=0.0020\,\text{m}^2$. Pressure is A. $2.0\times 10^5\,\text{Pa}$ B. $0.80\,\text{Pa}$ C. $400.002\,\text{Pa}$ D. $8.0\times 10^{-6}\,\text{Pa}$

$p=F/A=400/0.0020=2.0\times 10^5\,\text{Pa}$. B multiplied instead of dividing.

3. $20\,\text{cm}^2$ in $\text{m}^2$ is A. $0.20\,\text{m}^2$ B. $2.0\times 10^{-6}\,\text{m}^2$ C. $0.0020\,\text{m}^2$ D. $20\,\text{m}^2$

Divide $\text{cm}^2$ by $10^4$. A divided by $100$ only.

4. Snowshoes reduce pressure on snow because they A. increase the person’s weight B. decrease the person’s weight C. decrease the contact area D. increase the contact area, so $p=F/A$ falls

Weight is almost unchanged. Area is what you change.

5. At a point in a liquid at rest, the pressure A. acts equally in all directions B. acts only downwards C. acts only towards the surface D. is zero unless the liquid is flowing

Specification 5.6.

Quiz 4 — $p=h\rho g$ and total pressure

1. In $p=h\rho g$, $h$ is A. the sloping distance along the side of the tank B. the width of the tank C. the vertical depth below the free surface D. the height of the laboratory bench

Width does not appear in the equation.

2. $h=4.0\,\text{m}$, $\rho=1000\,\text{kg/m}^3$, $g=10\,\text{N/kg}$. Pressure difference is A. $4.0\times 10^4\,\text{Pa}$ B. $400\,\text{Pa}$ C. $1.4\times 10^5\,\text{Pa}$ D. $2.5\times 10^{-4}\,\text{Pa}$

$4.0\times 1000\times 10=4.0\times 10^4\,\text{Pa}$. C added an atmosphere that was not asked for.

3. Seawater $\rho=1020\,\text{kg/m}^3$, $h=2.5\,\text{m}$. Pressure due to the seawater is A. $255\,\text{Pa}$ B. $4080\,\text{Pa}$ C. $1.00\times 10^5\,\text{Pa}$ D. $2.55\times 10^4\,\text{Pa}$

$2.5\times 1020\times 10=25500\,\text{Pa}$.

4. For the seawater in Q3, atmosphere $=1.00\times 10^5\,\text{Pa}$. Total pressure at that depth is A. $2.55\times 10^4\,\text{Pa}$ B. $1.255\times 10^5\,\text{Pa}$ C. $7.45\times 10^4\,\text{Pa}$ D. $2.55\times 10^9\,\text{Pa}$

Add: $2.55\times 10^4+1.00\times 10^5=1.255\times 10^5\,\text{Pa}$. A forgot the atmosphere.

5. Two tanks hold the same liquid. Tank A is narrow, tank B is wide. At $1.0\,\text{m}$ depth the pressure due to the liquid is A. the same in both, because $p=h\rho g$ does not depend on width B. larger in B because there is more liquid C. larger in A because the liquid is squeezed D. zero in both unless the tanks are sealed

Same $h$, same $\rho$, same $g$ → same $p$.

Quiz 5 — Kelvin, kinetic theory, Boyle and $p/T$

1. Absolute zero is A. $0^\circ\text{C}$ B. $273^\circ\text{C}$ C. $-273^\circ\text{C}$ D. $100\,\text{K}$

Specification 5.16. Also $0\,\text{K}$. Particles have (almost) zero KE.

2. $20^\circ\text{C}$ in kelvin is A. $293\,\text{K}$ B. $20\,\text{K}$ C. $-253\,\text{K}$ D. $273\,\text{K}$

$T=\theta+273=293\,\text{K}$. D is $0^\circ\text{C}$.

3. A gas is compressed at constant temperature. The pressure rises because the particles A. move faster B. collide with the walls more frequently C. stop colliding D. lose all their kinetic energy

Constant $T$ means average speed is unchanged. Smaller $V$ → more frequent hits (5.20).

4. $p_1=100\,\text{kPa}$, $V_1=40\,\text{cm}^3$, $V_2=20\,\text{cm}^3$, constant $T$. $p_2$ is A. $50\,\text{kPa}$ B. $80\,\text{kPa}$ C. $140\,\text{kPa}$ D. $200\,\text{kPa}$

$p_1V_1=p_2V_2$ → $p_2=200\,\text{kPa}$. A inverted the volume ratio.

5. Gas at $120\,\text{kPa}$ and $20^\circ\text{C}$ is heated to $100^\circ\text{C}$ at constant volume. $p_2$ is about A. $600\,\text{kPa}$ B. $153\,\text{kPa}$ C. $24\,\text{kPa}$ D. $120\,\text{kPa}$

$T_1=293\,\text{K}$, $T_2=373\,\text{K}$, $p_2=120\times 373/293=153\,\text{kPa}$. A used °C in the ratio.