Density and Pressure — Question 7

Section 14-2 · Physics pages 2084–2085

Question 7 Density changes can reveal a leak during compression A gas sample initially has volume 2.0 L and density 1.5 kg/m3. After a process its volume is 0.80 L and density is measured as 3.0 kg/m3. Do not assume a gas equation of state. Compression changes volume; measured density and volume together determine whether mass was conserved. (a) Find the initial and final masses and the mass fraction lost. (b) What density would the final sample have if compression alone occurred in a sealed system? Can pressure be inferred from the given density data alone?
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Question 7 - Solution Density changes can reveal a leak during compression (a) Derivation and calculation mi = ρiVi = (1.50)(2.0 × 10−3) = .00300 kg. mf = ρfVf = (3.00)(.80 × 10−3) = .00240 kg. ∆mlost = .600 g , mi −mf mi = .200 = 20.0% . (b) Result and interpretation If the sample had remained sealed, ρf,sealed = mi Vf = .00300 .00080 = 3.75 kg/m3 . ρf,measured < ρf,sealed ⇒mass loss under the stated measurements. Density alone cannot determine pressure without an equation of state and additional thermodynamic information. Physical check Compression alone conserves mass; the observed decrease in ρV is the evidence for loss.

Question 7

Density changes can reveal a leak during compression

A gas sample initially has volume 2.0L2.0\,\mathrm{L} and density 1.5kg/m31.5\,\mathrm{kg/m^3}. After a process its volume is 0.80L0.80\,\mathrm{L} and density is measured as 3.0kg/m33.0\,\mathrm{kg/m^3}. Do not assume a gas equation of state.

Diagram shown in the original SVG worksheet.

  1. Find the initial and final masses and the mass fraction lost.

  2. What density would the final sample have if compression alone occurred in a sealed system? Can pressure be inferred from the given density data alone?

Question 7 - Solution

Density changes can reveal a leak during compression

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

mi=ρiVi=(1.50)(2.0×10−3)=.00300kg.m_i=\rho_iV_i=(1.50)(2.0\times10^{-3})=.00300\,\mathrm{kg}.mf=ρfVf=(3.00)(.80×10−3)=.00240kg.m_f=\rho_fV_f=(3.00)(.80\times10^{-3})=.00240\,\mathrm{kg}.Δmlost=.600g,mi−mfmi=.200=20.0%.\boxed{\Delta m_{\mathrm{lost}}=.600\,\mathrm g},\qquad\boxed{\frac{m_i-m_f}{m_i}=.200=20.0\%}.

(b) Result and interpretation

If the sample had remained sealed,ρf,sealed=miVf=.00300.00080=3.75kg/m3.\boxed{\rho_{f,\mathrm{sealed}}=\frac{m_i}{V_f}=\frac{.00300}{.00080}=3.75\,\mathrm{kg/m^3}}.ρf,measured<ρf,sealed⇒mass loss under the stated measurements.\rho_{f,\mathrm{measured}}<\rho_{f,\mathrm{sealed}}\Rightarrow\text{mass loss under the stated measurements}.Density alone cannot determine pressure without an equation of state and additional thermodynamic information.

Physical check

Compression alone conserves mass; the observed decrease in ρV\rho V is the evidence for loss.