Hydrogen Squeezed Under Sand

Q. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. Taking the ratio of specific heats of hydrogen as 1.4, by what factor does the pressure of the gas increase if the gas is compressed to half its original volume? expects 2.64

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Q. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. Taking the ratio of specific heats of hydrogen as 1.4 $γ = 1.4$ , by what factor does the pressure of the gas increase if the gas is compressed to half its original volume $V_{1} = \frac{V_{0}}{2}$ ?

The playground

Inputs

gamma

γ

1.4

Outputs

factor?

The solution

Solved with AdiabaticProcess

Given

$γ = 1.4$
$V_{1} = \frac{V_{0}}{2}$ from “compressed to half its original volume”

Find $\frac{P_{1}}{P_{0}}$ .

Working

pv to the gamma is kept(compressed) $P_{0} \cdot {V_{0}}^{γ} = P_{1} \cdot {V_{1}}^{γ}$
$P_{1} = \frac{P_{0} \cdot {V_{0}}^{1.4}}{{(\frac{V_{0}}{2})}^{1.4}}$
$\frac{P_{1}}{P_{0}} = 2.63902$

Conclusion

With no heat let in or out, squeezing the gas also warms it, so its pressure rises by more than the factor of two the volume alone would give.

The Ruby

1require_relative "../physics"23question :hydrogen_squeezed_under_sand do4  kind :numerical5  describes "A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The " \6            "walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of " \7            "sand on it. Taking the ratio of specific heats of hydrogen as %{gamma}, %{factor}?"8  given gamma: "1.4"9  asks factor: "by what factor does the pressure of the gas increase if the gas is compressed to half its original volume"10end

1require_relative "../questions/hydrogen_squeezed_under_sand"2require_relative "../science/thermodynamics/adiabatic_process"34solve :hydrogen_squeezed_under_sand, using: AdiabaticProcess do5  objects :gas6  gas.goes(:compressed)78  take gamma: gas.ratio_of_specific_heats910  deduce_from "compressed to half its original volume", gas.final_volume => gas.initial_volume / 21112  answer find_relation(gas.final_pressure / gas.initial_pressure), figures: 31314  conclusion "With no heat let in or out, squeezing the gas also warms it, so its pressure rises by more " \15             "than the factor of two the volume alone would give."16end

1require_relative "../solutions/hydrogen_squeezed_under_sand"23play_with :hydrogen_squeezed_under_sand do4  title "Hydrogen Squeezed Under Sand"5end

The laws

quantity	written	units	within	from
pressure	$P$	Pa	0 … ∞	AdiabaticProcess
volume	$V$	m³	0 … ∞	AdiabaticProcess
ratio of specific heats	$γ$	—	—	AdiabaticProcess
starting pressure	$P_{1}$	Pa	0 … ∞	AdiabaticProcess
final pressure	$P_{2}$	Pa	0 … ∞	AdiabaticProcess
starting volume	$V_{1}$	m³	0 … ∞	AdiabaticProcess
final volume	$V_{2}$	m³	0 … ∞	AdiabaticProcess
pressure of gas at start of compressed	$P_{0}$	Pa	0 … ∞
volume of gas at start of compressed	$V_{0}$	m³	0 … ∞
ratio of specific heats of gas	$γ$	—	—
pressure of gas at end of compressed	$P_{1}$	Pa	0 … ∞
volume of gas at end of compressed	$V_{1}$	m³	0 … ∞

P_{1} \equiv P_{i} (start of motion)

starting pressure

P_{2} \equiv P_{i} (end of motion)

final pressure

V_{1} \equiv V_{i} (start of motion)

starting volume

V_{2} \equiv V_{i} (end of motion)

final volume

P_{0} \cdot {V_{0}}^{γ} = P_{1} \cdot {V_{1}}^{γ}

pv to the gamma is kept(compressed)

V_{1} = \frac{V_{0}}{2}

compressed to half its original volume