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Physics Chapter 8 : Topic 6
Work Done by a Gas
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The work done when a gas expands at constant pressure is given by:
Reasoning: For constant pressure, work done by a gas is the product of pressure and change in volume.
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In the equation $\mathbf{W = p\mathrm{\Delta}V}$, the symbol $\mathbf{\mathrm{\Delta}V}$ represents:
Reasoning: $\mathbf{\mathrm{\Delta}V =}\mathbf{V}_{\mathbf{f}}\mathbf{-}\mathbf{V}_{\mathbf{i}}\mathbf{\ }$, which represents the change in volume.
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The SI unit of work done by a gas is:
Reasoning: Work is measured in joules $\mathbf{(}$J$\mathbf{)}$.
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The SI unit of pressure in $\mathbf{W = p\mathrm{\Delta}V}$ is:
Reasoning: Pressure must be in pascals for $\mathbf{SI}$ calculations.
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A gas expands against a constant pressure. The work done by the gas is:
Reasoning: During expansion, $\mathbf{\mathrm{\Delta}V}$ is positive, so $\mathbf{W = p\mathrm{\Delta}V}$ is positive.
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If a gas is compressed at constant pressure, the work done by the gas is:
Reasoning: Compression gives negative $\mathbf{\mathrm{\Delta}V}$, so work done by the gas is negative.
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A gas expands from $\mathbf{2\ m}^{\mathbf{3}}$ to $\mathbf{5\ m}^{\mathbf{3}}$ at a pressure of $\mathbf{1000\ Pa}$. The work done is:
Reasoning: $\mathbf{W = p\mathrm{\Delta}V = 1000\ (5 - 2) = 30}\mathbf{00\ J}$
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A gas expands by $\mathbf{0.02\ m}^{\mathbf{3}}$ against a pressure of $\mathbf{5 \times}\mathbf{10}^{\mathbf{4}}\mathbf{\ Pa}$. Work done is:
Reasoning: $\mathbf{W = p\mathrm{\Delta}V = (5 \times}\mathbf{10}^{\mathbf{4}}\mathbf{)(0.02) = 1000\ J}$
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If there is no change in volume of a gas, the work done is:
Reasoning: $\mathbf{W = p\mathrm{\Delta}V}$ ; If $\mathbf{\mathrm{\Delta}V = 0}$, then $\mathbf{W = 0}$.
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The work done by a gas during expansion represents:
Reasoning: Expansion allows gas to transfer energy by doing mechanical work.
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The area under a pressure-volume graph represents:
Reasoning: The area under a $\mathbf{(P - V)}$ graph equals work done.
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On a pressure-volume graph, constant pressure is represented by:
Reasoning: Constant pressure means pressure remains unchanged while volume varies.
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For a gas expanding at constant pressure, increasing the volume change will:
Reasoning: From $\mathbf{W = p\mathrm{\Delta}V}$, work is directly proportional to volume change.
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If pressure is doubled while volume change remains constant, work done:
Reasoning: $\mathbf{W \propto p}$, so doubling pressure doubles work.
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A gas expands from $\mathbf{1\ m}^{\mathbf{3}}$ to $\mathbf{4\ m}^{\mathbf{3}}$ at $\mathbf{(2000\ Pa)}$. The work done is:
Reasoning: $\mathbf{W = p\mathrm{\Delta}V = 2000(4 - 1) = 6000\ J}$
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Which condition is necessary for using $\mathbf{W = p\mathrm{\Delta}V}$?
Reasoning: The equation applies when pressure remains constant.
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During expansion of a gas, the surroundings:
Reasoning: The gas pushes surroundings and transfers energy as work.
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A gas does $\mathbf{500\ J}$ of work while expanding at constant pressure of $\mathbf{2500\ Pa}$. The change in volume is:
Reasoning: $\mathbf{\mathrm{\Delta}V =}$ $\frac{\mathbf{W}}{\mathbf{p}}\mathbf{=}\frac{\mathbf{500}}{\mathbf{2500}}\mathbf{= 0.2\ m}^{\mathbf{3}}$
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If a gas is compressed, the final volume is:
Reasoning: Compression reduces the volume.
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The equation $\mathbf{W = p\mathrm{\Delta}V}$ is derived from:
Reasoning: Work is force multiplied by distance, and pressure relates force to area.
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Pressure is defined as:
Reasoning: $\mathbf{p =}$ $\frac{\mathbf{F}}{\mathbf{A}}$
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A gas expands from $\mathbf{3L}$ to $\mathbf{8L}$ at $\mathbf{2 \times}\mathbf{10}^{\mathbf{5\ }}\mathbf{Pa}$. The work done is:
Reasoning: $\mathbf{\mathrm{\Delta}V = 8 - 3 = 5L = 5 \times}\mathbf{10}^{\mathbf{- 3}}\mathbf{\ m}^{\mathbf{3}}$$\mathbf{W = p\mathrm{\Delta}V}\mathbf{= (2 \times}\mathbf{10}^{\mathbf{5}}\mathbf{)\ (5 \times}\mathbf{10}^{\mathbf{- 3}}\mathbf{) = 1}\mathbf{000\ J}$
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Work done by a gas depends on:
Reasoning: According to $\mathbf{W = p\mathrm{\Delta}V}$, work depends on pressure and volume change.
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A gas expands at constant pressure. Which statement is correct?
Reasoning: Expansion means the gas performs work on surroundings.
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If pressure is constant and volume increases three times more than before, work done becomes:
Reasoning: Work is directly proportional to volume change.
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A piston moves outward because gas pressure pushes it. The gas is:
Reasoning: Outward piston movement means gas volume increases.
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The sign convention for work done by gas during expansion is usually:
Reasoning: Expansion means the gas transfers energy to surroundings.
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A gas has pressure $\mathbf{4 \times}\mathbf{10}^{\mathbf{4}}\mathbf{\ Pa}$ and expands by $\mathbf{0.5\ m}^{\mathbf{3}}$. Work done equals:
Reasoning: $\mathbf{W = p\mathrm{\Delta}V = (4 \times}\mathbf{10}^{\mathbf{4}}\mathbf{)(0.5) = 2 \times}\mathbf{10}^{\mathbf{4}}\mathbf{\ J}$
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If a gas expands at zero pressure, work done is:
Reasoning: $\mathbf{W = p\mathrm{\Delta}V}$ ; If $\mathbf{p = 0}$, work is zero.
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The physical meaning of $\mathbf{p\mathrm{\Delta}V}$ is:
Reasoning: Pressure acting through a volume change represents mechanical work.
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A gas expands at constant pressure from $\mathbf{10\ m}^{\mathbf{3}}$ to $\mathbf{12\ m}^{\mathbf{3}}$. If $\mathbf{p = 500\ Pa}$, work done is:
Reasoning: $\mathbf{W = p\mathrm{\Delta}V\ }\mathbf{= 500\ (12 - 10) = 1000\ J}$
32 / 40
Which graph gives the work done during constant pressure expansion?
Reasoning: Work is represented by the area under a $\mathbf{P - V}$ graph.
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A negative value of work done by gas indicates:
Reasoning: Com
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The equation $\mathbf{W = p\mathrm{\Delta}V}$ is an application of:
Reasoning: Work represents energy transfer.
35 / 40
A gas absorbs heat and expands at constant pressure. Some energy is used for:
Reasoning: Expanding gas performs work on surroundings.
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The unit conversion needed when volume is given in liters is:
Reasoning: $\mathbf{SI}$ calculations require volume in cubic meters.
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The work done by gas at constant pressure is maximum when:
Reasoning: $\mathbf{W = p\mathrm{\Delta}V}$, so larger values produce larger work.
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The work done by a gas increases if:
Reasoning: Work depends on the product of pressure and volume change.
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If $\mathbf{W = 1000\ J}$ and $\mathbf{\mathrm{\Delta}V =}\mathbf{0.5\ m}^{\mathbf{3}}$, pressure is:
Reasoning: $\mathbf{p =}$ $\frac{\mathbf{W}}{\mathbf{\mathrm{\Delta}V}}\mathbf{=}\frac{\mathbf{1000}}{\mathbf{0.5}}\mathbf{= 2000\ Pa}$
40 / 40
The equation $\mathbf{W = p\mathrm{\Delta}V}$ shows that work done is:
Reasoning: Work depends on both pressure and change in volume.
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