IB Chemistry S1.5: Ideal gases worked solutions

The ideal description neglects forces between particles between encounters. An appreciable attractive influence therefore contradicts that description, selecting A. A spread of speeds and directions is allowed. An elastic collision can exchange energy between particles while conserving their combined kinetic energy. [R: assumptions of the model.]

Elastic specifies conservation for the pair: add the kinetic energies of both particles before and after the encounter. The totals agree, but either particle may gain energy at the other's expense. B allows exactly this. [R: meaning of an elastic collision.]

Choose A. The higher temperature raises typical kinetic energy, reducing the importance of attractions relative to that energy. The lower pressure gives a more dilute sample, making the occupied particle volume less significant. B retains severe crowding, C retains low temperature, and D has both unfavourable conditions. [R: limits of the ideal approximation.]

Fixing both n and V in pV=nRT [B, booklet section 1] gives p_2/p_1=T_2/T_1 [D]. Calculate p_2=(8.40\times10^4)(350/280)=1.05\times10^5\,\mathrm{Pa}. Heating raises the result, so C has the correct size and direction.

The final volume is 0.80V_1. With the same n,T, p_1V_1=p_2V_2 [D from pV=nRT, B, section 1]. Thus p_2/p_1=1/0.80=1.25, giving C. A percentage change in volume cannot be added directly to the pressure percentage.

For an ideal gas V=nRT/p [D from the supplied gas equation, B, section 1]. Both samples have equal n,p,T, so their volumes agree: each occupies 2.27\times10^{-2}\,\mathrm{m^3} using the supplied STP molar volume [B, section 2]. One mole of oxygen molecules contains two moles of atoms; one mole of neon particles contains one mole of atoms. D satisfies both comparisons.

Use pV=nRT [B, section 1] and isolate n=pV/(RT) [D]. All inputs are already SI. The product pV=(9.60\times10^4)(4.20\times10^{-4})=40.32\,\mathrm J. Dividing by RT=8.31(280)\,\mathrm{J\,mol^{-1}} gives n=0.0173285\ldots\,\mathrm{mol}, which rounds to 0.0173\,\mathrm{mol}. Choose B.

Write p=(nR/V)T [D from pV=nRT, B, section 1]. Its coefficient of T is the gradient. Keeping V fixed while doubling n doubles that coefficient. There is no constant term to shift the ideal kelvin intercept, so B describes the new line.

(a) The volume and supplied molar volume refer to the same conditions. Divide n=V/V_m [D, supplied V_m in booklet section 2]: n=(3.405\times10^{-4})/(2.27\times10^{-2})=0.0150\,\mathrm{mol} [1]. (b) Each mole contains N_{\mathrm A} particles [B, section 2], so N=nN_{\mathrm A}=0.0150(6.02\times10^{23})=9.03\times10^{21} particles [D, 1].

(a) The rigid sealed vessel keeps n,V fixed, so p_2=p_1T_2/T_1 [1, D from pV=nRT, B, section 1]. Substitute p_2=(9.00\times10^4)(400/320)=112500\,\mathrm{Pa}, or 1.13\times10^5\,\mathrm{Pa} to three significant figures [1]. (b) Increasing temperature raises mean kinetic energy and typical particle speed [1]. The particles reach the walls more frequently and deliver more momentum at each impact; the resulting larger wall force per area raises pressure [1]. These are linked physical statements, rather than a claim that the particles grow.

(a) Convert to T_1=12.0+273.15=285.15\,\mathrm K and T_2=58.0+273.15=331.15\,\mathrm K [1; B, section 4]. For unchanged amount, the supplied combined law [B, section 1] rearranges to p_2=p_1V_1T_2/(T_1V_2) [1, D]. Enter the SI data: p_2=(1.08\times10^5)(7.50\times10^{-4})(331.15)/[(285.15)(1.10\times10^{-3})]=85515.279\ldots\,\mathrm{Pa}. Hence p_2=8.55\times10^4\,\mathrm{Pa} [1]. (b) The temperature ratio is about 1.16, whereas the volume ratio is about 1.47. Expansion therefore more than offsets the heating effect, leaving a net pressure fall [1].

(a) Divide by the supplied STP molar volume [B, section 2]: n=V/V_m=(9.08\times10^{-4})/(2.27\times10^{-2})=0.0400\,\mathrm{mol} [1, D]. Count particles with N=nN_{\mathrm A} [D], using N_{\mathrm A} from section 2: N=0.0400(6.02\times10^{23})=2.408\times10^{22}, or 2.41\times10^{22} particles [1]. (b) Booklet STP uses 273.15\,\mathrm K and 1.00\times10^5\,\mathrm{Pa} [1; section 4]. (c) Dividing pV=nRT [B, section 1] gives V_m=RT/p [D]. No chemical-identity variable remains, so another ideal gas at the same conditions has the same molar volume [1].

(a) Conservation of the pair's kinetic energy says nothing about whether the particles take up appreciable space [1]. At extreme compression their finite volume occupies a substantial share of the container, defeating the separate point-particle assumption [1]. (b) Cooling lowers typical kinetic energy [1]. Attractions can then influence motion more strongly relative to that energy, so ideal behaviour need not improve [1]. The substance remaining gaseous does not restore either approximation.

(a) Draw a smooth decreasing curve through the overall scatter trend [1]; do not join successive points with straight segments. Fixed n,T in pV=nRT [B, booklet section 1] gives p\propto1/V [D, 1]. (b) The scale factors make X represent p=1.60\times10^5\,\mathrm{Pa} and V=3.00\times10^{-4}\,\mathrm{m^3} [1]. Use n=pV/(RT) [D]: n=48.0/[8.31(310)]=0.0186328170\ldots\,\mathrm{mol}=0.0186\,\mathrm{mol} [1]. The plotted values are synthetic readings, not an IB dataset.

(a) Convert p=9.80\times10^4\,\mathrm{Pa} and V=1.20\times10^{-2}\,\mathrm{m^3} [1]. With pV=nRT [B, section 1], n=1176/[8.31(295)]=0.479716086\ldots\,\mathrm{mol}=0.480\,\mathrm{mol} [1]. (b) At matching pressure and temperature, V=nRT/p makes volume proportional to amount [D, 1]. Thus x=1.80/100=0.0180 and n(\ce{CO2})=0.0180(0.479716086\ldots)=0.00863488956\ldots\,\mathrm{mol}, or 8.63\times10^{-3}\,\mathrm{mol} [1].

(a) The booklet relation n=m/M [B, section 1] gives n=0.640/32.00=0.0200\,\mathrm{mol} [1]. Convert V=1.60\times10^{-3}\,\mathrm{m^3} and rearrange pV=nRT [B, section 1] to p=nRT/V [D, 1]. Calculate p=0.0200(8.31)(305)/(1.60\times10^{-3})=31681.875\,\mathrm{Pa}=31.7\,\mathrm{kPa} [1]. (b) Helium has 0.640/4.00=0.160\,\mathrm{mol}, eight times the amount. At fixed V,T, p\propto n, so its pressure is eight times greater [1].

(a) For run I, V=(nR/p)T [D from pV=nRT, B, section 1]. Since T/\mathrm K=t/{}^\circ\mathrm C+273.15 [B, section 4], draw an increasing straight line with positive V at zero Celsius and extrapolated intercept -273.15\,{}^\circ\mathrm C [1]. A real sample may condense before that limit; zero-volume ideal behaviour is not a reachable gas state [1]. (b) Run II has n=pV/(RT) [D], an inverse curve in the positive quadrant, not a falling straight line [1]. The amount ratio is n_2/n_1=250/400=0.625; the decrease is 37.5\% [1]. The valve means that fixed amount cannot be assumed.

Use the supplied pV=nRT and n=m/M [B, section 1] with density \rho=m/V [R]. Eliminating m,n,V gives M=\rho RT/p [D]. With SI density, M=3.20(8.31)(315)/(1.40\times10^5)=0.059832\,\mathrm{kg\,mol^{-1}}. Multiply by 1000 to obtain 59.8\,\mathrm{g\,mol^{-1}} to three significant figures, selecting C.

(a) Rearrange pV=nRT [B, section 1] as n=pV/(RT) [1, D]. The numerator is 19.08\,\mathrm J and the denominator is 3240.9\,\mathrm{J\,mol^{-1}}, giving n=0.005887253\ldots\,\mathrm{mol}=0.00589\,\mathrm{mol} [1]. From n=m/M [B, section 1], M=m/n [D]; retain unrounded n to obtain M=69.981698\ldots\,\mathrm{g\,mol^{-1}}=70.0\,\mathrm{g\,mol^{-1}} [1]. (b) One empirical unit has mass 12.01+2(1.01)=14.03. The multiplier is 69.981698\ldots/14.03=4.988\ldots, consistent with 5 [1]. Multiplying both subscripts gives \ce{C5H10} [1]. (c) Retained air contributes pressure that is incorrectly attributed to the added sample, inflating the calculated amount [1]. The same mass gain divided by this inflated amount gives too small a molar mass [1]. Here the residual air is retained in both weighings, so it is not part of the added mass.

(a) One possible pair is negligible particle size compared with the container [1] and no attractions between particles between encounters [1]. An elastic-collision assumption can replace either one. (b) Choose X [1]. With attractions dominant, its stronger inward pulls reduce the momentum delivered to the wall more than for Y, producing the larger downward departure [1]. The comparison holds n,V,T fixed. (c) Compression reduces the average separation [1]. The unchanged finite particle size then occupies an appreciable share of the available volume, so a point-particle treatment fails [1]. [R: model assumptions; no quantitative real-gas correction is required.]

(a) Apply pV=nRT [B, section 1] separately because gas is removed. In SI, n_1=120000(0.00250)/[8.31(300)]=0.120336943\ldots\,\mathrm{mol} [1] and n_2=90000(0.00300)/[8.31(360)]=0.0902527076\ldots\,\mathrm{mol} [1]. The initial fraction removed is (n_1-n_2)/n_1=0.250, hence 25.0\% carbon dioxide [1]. (b) The removed amount is 0.0300842359\ldots\,\mathrm{mol} [1]. Use m=nM [D from booklet n=m/M, B, section 1]: the gain is 0.0300842359\ldots(44.01)=1.32400722\ldots\,\mathrm g=1.32\,\mathrm g [1]. (c) Cancelling n is invalid because the absorber changes it [1]. The fixed-amount expression would give p_2=120(2.50/3.00)(360/300)=120\,\mathrm{kPa} [1], above the measured 90.0\,\mathrm{kPa}.

(a) Balance \ce{Mg + 2HCl -> MgCl2 + H2} [1]. Convert V=2.55\times10^{-4}\,\mathrm{m^3} and p=1.01\times10^5\,\mathrm{Pa} [1]. From pV=nRT [B, section 1], n(\ce{H2})=25.755/[8.31(298)]=0.0104002617\ldots\,\mathrm{mol} [1]. (b) The balanced mole ratio is 1:1, so m(\ce{Mg})=0.0104002617\ldots(24.31)=0.252830361\ldots\,\mathrm g [1]. Purity is 100(0.252830361\ldots)/0.280=90.2965576\ldots\%=90.3\% [1]. (c) A leak lowers collected gas amount, which lowers inferred magnesium mass and purity [1].

(a) The first interval produces 18.0\times10^{-6}\,\mathrm{m^3}. With pV=nRT [B, section 1], \Delta n=100000(18.0\times10^{-6})/[8.31(300)]=7.22021661\times10^{-4}\,\mathrm{mol} [1]. Divide by 20\,\mathrm s: 3.61010830\times10^{-5}\,\mathrm{mol\,s^{-1}}, or 3.61\times10^{-5} [1]. (b) The next interval gains 11.0\,\mathrm{cm^3} in the same time, so its mean rate is 11/18=0.611 times the first [1], or 2.21\times10^{-5}\,\mathrm{mol\,s^{-1}}. Reactant depletion can lower the frequency of successful encounters [1]; these observations alone do not establish a rate law. (c) Higher temperature increases volume per mole as well as potentially changing reaction rate [1]. Convert each volume increment using its own T: \Delta n/\Delta t=p\Delta V/(RT\Delta t) [D, 1].

(a) V=nRT/p [D from pV=nRT, B, section 1] gives equal ideal volumes because n,p,T agree [1]. Equal temperature means equal mean translational kinetic energy [1]; molecular mass need not agree. (b) Ammonia can form hydrogen bonds between molecules, while methane has dispersion forces only [1]. The stronger attractions in ammonia become more significant relative to kinetic energy on cooling and can produce a greater departure [1]; this is qualitative, not a numerical equation of state. (c) The point-particle model neglects occupied molecular volume [1]. When compressed, particle volume becomes a substantial fraction of the total, which can make the real volume larger than the ideal prediction [1]. Attractions alone do not explain this direction.