A bicycle tyre has a volume of $1.20 \times 10^{-3} \mathrm{~m}^{3}$ when fully inflated. The barrel of the bicycle pump has a working volume of $9.0 \times 10^{-5} \mathrm{~m}^{3}$. How many strokes of the pump are needed to completely inflate the flat tyre to a total pressure of $3.0 \times 10^{5} \mathrm{~Pa}$ ? The atmospheric pressure is $1.00 \times 10^{5} \mathrm{~Pa}$. Assume the air is pumped in slowly, so that the temperature remains constant. [4]
A bicycle tyre has a volume of $1.2 \times 10^{-3} \mathrm{~m}^{3}$ when fully inflated. A bicycle pump has a working volume of $9.0 \times 10^{-5} \mathrm{~m}^{3}$. How many strokes, $n$, of the pump are needed to inflate the completely flat tyre, containing no air, to a pressure of $3.0 \times 10^{5} \mathrm{~Pa}$ ? The atmospheric pressure is $1.0 \times 10^{5} \mathrm{~Pa}$. Assume the air is pumped in slowly so that there is no change in temperature. [5]
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Two identical spherical glass containers are joined by a narrow tube, whose volume is negligible compared to the spheres. The spheres contain air at $100^{\circ} \mathrm{C}$. One of the spheres is then heated by $50^{\circ} \mathrm{C}$ whilst the other is cooled by $50^{\circ} \mathrm{C}$. This produces a small change in pressure, from $P_{\text {initial }}$ to $P_{\text {final }}$, of the air in the system. What common temperature of the two spheres could produce the same final pressure $P_{\text {final }}$ ? [4]
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- The common temperature is $93.3^\circ C$, slightly below the original $100^\circ C$.
Worked solutions are in the book.
Two glass bulbs are connected by a thin tube. One glass bulb has a volume of $75 \mathrm{~cm}^{3}$, the other $150 \mathrm{~cm}^{3}$, and gas can move freely between them. Initially the system contains nitrogen at $-12^{\circ} \mathrm{C}$ and $0.91 \times 10^{5} \mathrm{~Pa}$. The smaller bulb is then warmed to $24^{\circ} \mathrm{C}$, whilst the larger bulb is maintained at $-12^{\circ} \mathrm{C}$.
Calculate the new pressure in the system. Assume the thermal expansion of the bulbs and the volume of the connecting tube are negligible. [5]
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- The new pressure is $9.48 \times 10^4 \, \text{Pa}$, slightly above the initial value.
Worked solutions are in the book.
A fixed mass of gas expands isothermally and the relationship between the pressure $p$, and the volume of the gas $V$, is $p V=380 \mathrm{~Pa} \mathrm{~m}^{3}$. The volume increases at a rate of $0.005 \mathrm{~m}^{3} \mathrm{~s}^{-1}$ when the volume is $0.17 \mathrm{~m}^{3}$. At what rate does the pressure decrease at this point? [3]
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A hot air balloon uses a gas flame to heat the air it contains to a temperature required to enable it to hover at a small distance above ground level. The mass of the balloon, ropes, basket and riders is $240 \mathrm{~kg}$ and the volume of the balloon is $1100 \mathrm{~m}^{3}$. The temperature of the surrounding air is $15^{\circ} \mathrm{C}$ and its density is $1.23 \mathrm{~kg} \mathrm{~m}^{-3}$.
To what temperature does the air in the balloon need to be heated? [5]
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- The air must be heated to about $350$ K ($77^\circ\text{C}$).
Worked solutions are in the book.
A gas is found to obey the equation relating $p, V, n, R, T$
$$ p(V-b)=n R T \exp \left(\frac{-a}{n R T V}\right) $$
where $p$ is the gas pressure
$V$ is its volume
$R$ is the molar gas constant
$n$ is the number of moles
$a$ and $b$ are constants.
(i) Determine the SI base units (m, kg, s) in which $a$ and $b$ are expressed.
(ii) If $b \ll V$ and $a \ll n R T V$, show that this expression approximates to the ideal gas equation relating $p, V, n, R, T$ at a particular temperature $T_{\mathrm{c}}$. Determine $T_{\mathrm{c}}$ in terms of $a, b, n$ and $R$.
Hint: For $x \ll 1 \quad \exp (x)=e^{x} \approx 1+x$ [4]
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- So $[a] = \text{kg} \cdot \text{m}^5 \cdot \text{s}^{-2}$ and $[b] = \text{m}^3$.
Worked solutions are in the book.
In a closed tube of length $l$ and cross-section area $A$ there is a volume of high pressure gas $P_1$ and an equal volume of low pressure gas $P_2$, separated by a diaphragm in the centre of the tube. The tube is uniformly heated from $T_0$ until the diaphragm bursts, which occurs when a maximum force $F$ is applied. After a moment the pressure equalises in the tube, which is now insulated from the surroundings so that no thermal energy is lost. What is the final pressure inside the tube? [6]
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- The final pressure is $P_\text{final} = \dfrac{F}{2A}\,\dfrac{P_1+P_2}{P_1-P_2}$; both $T_0$ and $V$ cancel.
Worked solutions are in the book.