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The ideal gas law is PV=nRTP V = n R T, where n=mMn = \frac{m}{M}, with mm being the sample mass in grams and MM the molar mass in g/mol\text{g/mol}. If pressure PP is measured in kilopascals (kPa\text{kPa}), volume VV in liters (L\text{L}), temperature TT in kelvins (K\text{K}), and the universal gas constant is specified as R∗=8.314×10−3 kPa⋅m3/(mol⋅K)R^* = 8.314 \times 10^{-3} \text{ kPa}\cdot\text{m}^3/(\text{mol}\cdot\text{K}) with conversion factor 1 m3=1000 L1 \text{ m}^3 = 1000 \text{ L}, express the mass density ρ=mV\rho = \frac{m}{V} explicitly in terms of PP, MM, R∗R^*, and TT without numerical approximation.

Answer and steps

Final answer

ρ=PM1000R∗T\rho = \frac{P M}{1000 R^* T}

Steps

01

Substitute the expression for the number of moles

To express the mass density ρ=mV\rho = \frac{m}{V} in terms of PP, MM, R∗R^*, and TT, we rearrange the ideal gas equation while accounting for unit consistency between liters and cubic meters.

The number of moles is given by n=mMn = \frac{m}{M}. Substituting this into the ideal gas law gives: PVconsistent=(mM)RTP V_{\text{consistent}} = \left(\frac{m}{M}\right) R T

02

Align the volume units with the gas constant

The constant R∗R^* is defined with volume in cubic meters (m3\text{m}^3), but volume VV is measured in liters (L\text{L}).

Since 1 m3=1000 L1\text{ m}^3 = 1000\text{ L}, a volume of V LV\text{ L} corresponds to: Vm3=V1000V_{\text{m}^3} = \frac{V}{1000}

Substituting Vm3V_{\text{m}^3} and R∗R^* into the gas law: P(V1000)=mMR∗TP \left(\frac{V}{1000}\right) = \frac{m}{M} R^* T

03

Rearrange the equation to isolate ρ=mV\rho = \frac{m}{V}

Multiply both sides by MM: PVM1000=mR∗T\frac{P V M}{1000} = m R^* T

Divide both sides by VV: PM1000=(mV)R∗T\frac{P M}{1000} = \left(\frac{m}{V}\right) R^* T

Since density is defined as ρ=mV\rho = \frac{m}{V}, replace mV\frac{m}{V} with ρ\rho: PM1000=ρR∗T\frac{P M}{1000} = \rho R^* T

Finally, isolate ρ\rho by dividing both sides by R∗TR^* T: ρ=PM1000R∗T\rho = \frac{P M}{1000 R^* T}

Common Pitfall: Forgetting the unit conversion factor when matching volume in liters to a gas constant given in m3\text{m}^3 will introduce a factor-of-10001000 error. Always ensure all base units in the formula match those defined in the constant.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student directly replaces R with R* without converting the volume unit of R* from cubic meters to liters, yielding an answer off by a factor of 1000.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    PV=nRT=(mM)RT  ⟹  ρ=mV=PMRTP V = n R T = \left(\frac{m}{M}\right) R T \implies \rho = \frac{m}{V} = \frac{P M}{R T}

  2. Step 2

    Using \text{Using } R = R^*   ⟹  \implies ρ\rho = PMR∗T\frac{P M}{R^* T}

    Correct

    R=1000R∗  ⟹  ρ=PM1000R∗TR = 1000 R^* \implies \rho = \frac{P M}{1000 R^* T}   or\text{or}   P(V1000)\left(\frac{V}{1000}\right) = mM\frac{m}{M} R^* T   ⟹  \implies ρ\rho = PM1000R∗T\frac{P M}{1000 R^* T}$

    The given constant R∗R^* is defined using cubic meters (m3\text{m}^3), whereas volume VV is measured in liters (L\text{L}). Because 1 m3=1000 L1\text{ m}^3 = 1000\text{ L}, the volume in the formula matching R∗R^* is V1000\frac{V}{1000}, meaning R=1000R∗R = 1000 R^*, introducing a factor of 10001000 into the denominator.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly substituted the molar mass relationship n=mMn = \frac{m}{M} into the ideal gas law.

You isolated the density expression ρ=mV\rho = \frac{m}{V} accurately using literal algebra.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Missing unit conversion factor

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WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You treated the specified constant R∗R^* as numerically and dimensionally interchangeable with the standard constant RR without checking that the volume unit in R∗R^* (m3\text{m}^3) differed from the given volume unit (L\text{L}).

REASONING PATH
Your ACEpath teacher’s suggested approach
Unit Alignment Check

  • Every derived literal quantity must match the base units defined inside dimensional constants.

  • Check the units of each given constant; if a parameter is in liters while the constant uses cubic meters, replace VV with V1000\frac{V}{1000} before solving.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Align the volume unit with R∗R^* by replacing VV with its cubic-meter equivalent: Vm3=V1000V_{\text{m}^3} = \frac{V}{1000}.

  2. Substitute into the gas law: P(V1000)=(mM)R∗TP\left(\frac{V}{1000}\right) = \left(\frac{m}{M}\right) R^* T.

  3. Isolate ρ=mV\rho = \frac{m}{V}: PM1000=(mV)R∗T=ρR∗T\frac{P M}{1000} = \left(\frac{m}{V}\right) R^* T = \rho R^* T.

  4. Divide by R∗TR^* T to find ρ\rho.

Final answer

ρ=PM1000R∗T\rho = \frac{P M}{1000 R^* T}

NEXT STEPS
Final Takeaways from Your Teacher

  • Write out the dimensions of each variable and constant before substituting them into a formula.

  • Practice converting literal equations where one variable's unit does not match the compound unit of a constant.

OverviewACEpath Teacher's Diagnosis

02 The student multiplies by 1000 instead of dividing by 1000 due to inverting the volume unit conversion when substituting R* into the ideal gas law.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    PV = mM\frac{m}{M} R T   ⟹  \implies ρ\rho = mV\frac{m}{V} = PMRT\frac{P M}{R T}

  2. Step 2

    1 m\text{ m}^3 = 1000 L\text{ L},  so \text{ so } R = R∗1000\frac{R^*}{1000}

    Correct

    1 m3=1000 L  ⟹  R=1000R∗ (or Vm3=V1000)1\text{ m}^3 = 1000\text{ L} \implies R = 1000 R^*   \left(\text{or } V_{\text{m}^3} = \frac{V}{1000}\right)

    Because 1 m3=1000 L1\text{ m}^3 = 1000\text{ L} sits in the numerator of the unit kPa⋅m3/(mol⋅K)\text{kPa}\cdot\text{m}^3/(\text{mol}\cdot\text{K}), substituting 1 m3=1000 L1\text{ m}^3 = 1000\text{ L} yields R=1000R∗R = 1000 R^*. Dividing R∗R^* by 10001000 inverts the conversion relationship.

  3. Step 3

    ρ\rho = PM(R∗/1000)T\frac{P M}{(R^*/1000) T}   ⟹  \implies ρ\rho = 1000PMR∗T\frac{1000 P M}{R^* T}

    Correct

    ρ\rho = PM(1000R∗)T\frac{P M}{(1000 R^*) T} = PM1000R∗T\frac{P M}{1000 R^* T}

    Your algebraic simplification from the substituted fraction was mathematically consistent, but it inherited the inverted factor of 10001000 from Step 2.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly substituted the mole definition n=mMn = \frac{m}{M} into the ideal gas equation.

You correctly isolated the density expression ρ=mV\rho = \frac{m}{V} in terms of the other variables.

You recognized that the volume unit difference between L\text{L} and m3\text{m}^3 required adjusting the formula.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Inverted unit conversion factor in literal equation

You divided R∗R^* by 10001000 instead of recognizing that 1 m31\text{ m}^3 equals 1000 L1000\text{ L} in the numerator of the unit definition.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing whether to multiply or divide when replacing a unit within a compound unit expression. A safe alternative is to convert the variable VV directly: since VV is in liters, the volume in cubic meters is V1000\frac{V}{1000}, leading directly to P(V1000)=mMR∗TP\left(\frac{V}{1000}\right) = \frac{m}{M}R^* T.

REASONING PATH
Your ACEpath teacher’s suggested approach
Unit Substitution Check

  • When converting compound units, substitute the equivalent quantity directly into the unit expression.

  • Write R∗=8.314×10−3kPa⋅(1 m3)mol⋅K=8.314×10−3kPa⋅(1000 L)mol⋅K=1000R∗kPa⋅Lmol⋅KR^* = 8.314 \times 10^{-3}\frac{\text{kPa}\cdot(1\text{ m}^3)}{\text{mol}\cdot\text{K}} = 8.314 \times 10^{-3}\frac{\text{kPa}\cdot(1000\text{ L})}{\text{mol}\cdot\text{K}} = 1000 R^*\frac{\text{kPa}\cdot\text{L}}{\text{mol}\cdot\text{K}}.

Variable Replacement Check

  • Convert the measured variable into standard formula units rather than modifying the defined constant.

  • Keep R∗R^* unchanged and substitute the volume in m3\text{m}^3 as Vm3=V1000V_{\text{m}^3} = \frac{V}{1000} directly into PVm3=nR∗TP V_{\text{m}^3} = n R^* T.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Express the volume in cubic meters matching the unit of R∗R^*: Vm3=V1000V_{\text{m}^3} = \frac{V}{1000}.

  2. Substitute Vm3V_{\text{m}^3} and n=mMn = \frac{m}{M} into the gas law: P(V1000)=mMR∗TP \left(\frac{V}{1000}\right) = \frac{m}{M} R^* T.

  3. Rearrange to solve for ρ=mV\rho = \frac{m}{V} by dividing both sides by VV and multiplying by MM: PM1000=(mV)R∗T\frac{P M}{1000} = \left(\frac{m}{V}\right) R^* T.

  4. Divide by R∗TR^* T to isolate ρ\rho.

Final answer

ρ=PM1000R∗T\rho = \frac{P M}{1000 R^* T}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice converting compound units by directly replacing unit symbols with their conversion equivalents.

  • Whenever possible, convert the measured physical variable (VV) rather than modifying a given physical constant (R∗R^*).

  • Perform a sanity check on unit size: a liter is much smaller than a cubic meter, so a fixed mass of gas occupies fewer cubic meters than liters, meaning Vm3=V/1000V_{\text{m}^3} = V/1000.

OverviewACEpath Teacher's Diagnosis

03 The student inverts the ratio of molar mass and pressure when rearranging the ideal gas law for density.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    PV=(mM)RTPV = \left(\frac{m}{M}\right)RT

  2. Step 2

    R = 1000 R^*

  3. Step 3

    ρ=mV=PMRT\rho = \frac{m}{V} = \frac{P}{MRT}

    Correct

    ρ=mV=PMRT\rho = \frac{m}{V} = \frac{PM}{RT}

    When rearranging PV=mMRTPV = \frac{m}{M}RT to solve for mV\frac{m}{V}, you divided by MM instead of multiplying by MM. Multiplying both sides by MM yields PVM=mRTPVM = mRT, which gives mV=PMRT\frac{m}{V} = \frac{PM}{RT}.

  4. Step 4

    ρ=P1000MR∗T\rho = \frac{P}{1000 M R^* T}

    Correct

    ρ=PM1000R∗T\rho = \frac{PM}{1000 R^* T}

    Substituting R=1000R∗R = 1000 R^* carried forward the misplaced MM from the previous step, placing it in the denominator of your final expression.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly substituted the mole relation n=mMn = \frac{m}{M} into the ideal gas equation.

You accurately identified the 1000×1000\times conversion relationship between volume in liters and the cubic meter base unit of R∗R^*.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Literal equation inversion error

You placed the molar mass MM in the denominator instead of multiplying it to the opposite numerator.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

When isolating a ratio of variables from an equation where one variable is in a denominator, you treated moving MM across the equals sign as dividing rather than multiplying.

REASONING PATH
Your ACEpath teacher’s suggested approach
Clear Denominators First

  • If a variable appears in a fraction's denominator, multiply both sides of the equation by that denominator to bring it to the numerator.

  • From PV=mRTMPV = \frac{mRT}{M}, multiply both sides by MM to get PVM=mRTPVM = mRT before isolating mV\frac{m}{V}.

Dimensional Sanity Check

  • Density ρ=massvolume\rho = \frac{\text{mass}}{\text{volume}} must increase when molar mass MM increases at fixed PP and TT.

  • Ensure MM is directly proportional to ρ\rho, meaning MM must sit in the numerator of the density formula.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Start from your equation: PV=(mM)RTPV = \left(\frac{m}{M}\right)RT

  2. Multiply both sides by MM to clear the fraction: PVM=mRTPVM = mRT

  3. Divide both sides by VV to isolate mV\frac{m}{V}: mV=PMRT\frac{m}{V} = \frac{PM}{RT}

  4. Substitute R=1000R∗R = 1000 R^* and ρ=mV\rho = \frac{m}{V}: ρ=PM1000R∗T\rho = \frac{PM}{1000 R^* T}

Final answer

ρ=PM1000R∗T\rho = \frac{PM}{1000 R^* T}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice isolating literal ratios by writing out each intermediate multiplication step explicitly instead of moving variables mentally.

  • Use physical intuition as a quick check: a gas with heavier molecules (larger MM) is denser at the same pressure and temperature, so MM must be in the numerator.

Lock it in

Question

When a formula uses a volume VV in liters (L\text{L}), how do you express that volume in cubic meters (m3\text{m}^3)?

Answer

Vm3=V1000V_{\text{m}^3} = \frac{V}{1000}, because 1 m3=1000 L1\text{ m}^3 = 1000\text{ L}.

Similar question

01

A flow formula is Q=AvQ = A v, where speed vv is in cm/s\text{cm/s} and area AA is in cm2\text{cm}^2. If you must use a constant velocity v∗=0.01vv^* = 0.01 v (where v∗v^* is in m/s\text{m/s}), express QQ explicitly in terms of AA and v∗v^*.

Answer

Since v=100v∗v = 100 v^*, substitute into Q=AvQ = A v to get Q=100Av∗Q = 100 A v^*. Key steps: 1 m/s=100 cm/s  ⟹  v=100v∗1\text{ m/s} = 100\text{ cm/s} \implies v = 100 v^*, then substitute directly into Q=A(100v∗)Q = A(100 v^*).

02

The hydrostatic pressure equation is P=ρghP = \rho g h, where density ρ\rho is in kg/m3\text{kg/m}^3, gravitational acceleration gg is in m/s2\text{m/s}^2, and depth hh is given in centimeters (cm\text{cm}). Given 1 m=100 cm1\text{ m} = 100\text{ cm}, express density ρ\rho explicitly in terms of PP, gg, and hh.

Answer

ρ=100Pgh\rho = \frac{100 P}{g h}. Key steps: depth in meters is hm=h100h_{\text{m}} = \frac{h}{100}, giving P=ρg(h100)=ρgh100P = \rho g \left(\frac{h}{100}\right) = \frac{\rho g h}{100}. Multiplying by 100100 and dividing by ghg h yields ρ=100Pgh\rho = \frac{100 P}{g h}.

03

Kinetic energy is K=12mv2K = \frac{1}{2} m v^2, where KK is in joules (J=kg⋅m2/s2\text{J} = \text{kg}\cdot\text{m}^2/\text{s}^2) and mass mm is measured in grams (g\text{g}). If speed is given as vv in m/s\text{m/s}, express the mass mm explicitly in terms of KK and vv using the conversion 1 kg=1000 g1\text{ kg} = 1000\text{ g}.

Answer

m=2000Kv2m = \frac{2000 K}{v^2}. Key steps: mass in kilograms is mkg=m1000m_{\text{kg}} = \frac{m}{1000}, so K=12(m1000)v2=mv22000K = \frac{1}{2}\left(\frac{m}{1000}\right)v^2 = \frac{m v^2}{2000}. Multiplying both sides by 20002000 and dividing by v2v^2 isolates mm.

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