Related questions with answers

The following table presents the number of reports of graffiti in each of New York’s five boroughs over a one-year period. These reports were classified as being open, closed, or pending.

OpenClosedPendingBoroughReportsReportsReportsTotalBronx1,1211,622802,823Brooklyn1,1702,706483,924Manhattan7443,380254,149Queens1,3532,043253,421Staten Island831180201Total4,4719,86917814,518\begin{array}{lrrrr} \hline & \text{Open} & \text{Closed} & \text{Pending} & \\ \text{Borough} & \text{Reports} & \text{Reports} & \text{Reports} & \text{Total}\\ \hline \text{Bronx} & 1,121 & 1,622 & 80 & 2,823\\ \text{Brooklyn} & 1,170 & 2,706 & 48 & 3,924\\ \text{Manhattan} & 744 & 3,380 & 25 & 4,149\\ \text{Queens} & 1,353 & 2,043 & 25 & 3,421\\ \text{Staten Island} & 83 & 118 & 0 & 201\\ \hline \text{Total} & 4,471 & 9,869 & 178 & 14,518\\ \end{array}

A graffiti report is selected at random. Compute the following probabilities.

a.\textbf{a.}\hspace{10pt} The report is open and comes from Brooklyn.

b.\textbf{b.}\hspace{10pt} The report is closed or comes from Queens.

c.\textbf{c.}\hspace{10pt} The report comes from Manhattan.

d.\textbf{d.}\hspace{10pt} The report does not come from Manhattan.

e.\textbf{e.}\hspace{10pt} The report is pending.

f.\textbf{f.}\hspace{10pt} The report is from the Bronx or Staten Island.

Question

Consider the transition between two forms of solid tin, Sn(s\operatorname{Sn}(s, gray )Sn(s) \rightarrow \operatorname{Sn}(s, white )). The two phases are in equilibrium at 1bar1 \mathrm{bar} and 18C18^{\circ} \mathrm{C}. The densities for gray and white tin are 57505750 and 7280 kg m37280 \mathrm{~kg} \mathrm{~m}^{-3}, respectively, and the molar entropies for gray and white tin are 44.1444.14 and 51.18 J K1 mol151.18 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, respectively. Calculate the temperature at which the two phases are in equilibrium at 350.bar350 . \mathrm{bar}.

Solution

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We are given the following data:

Ti=18C=18+273.15=291.15KPi=1bar=1×105PaρSn,gray=5750kgm3ρSn,white=7280kgm3SSn,gray=44.14JK.molSSn,white=51.18JK.molPf=350bar=350×105Pa\begin{aligned} &T_i= 18^{\circ} C\\ &=18+273.15=291.15\:\text{K}\\ \\ &P_i= 1\:\text{bar}=1\times10^5\:\text{Pa}\\ \\ &\rho_{Sn,gray}= 5750\:\frac{\text{kg}}{\text{m}^3}\\ \\ &\rho_{Sn,white}= 7280\:\frac{\text{kg}}{\text{m}^3}\\ \\ &S_{Sn,gray}= 44.14\:\frac{\text{J}}{\text{K.mol}}\\ \\ &S_{Sn,white}= 51.18\:\frac{\text{J}}{\text{K.mol}}\\ \\ &P_f=350\:\text{bar}= 350 \times 10^5\:\text{Pa}\\ \\ \end{aligned}

Required:

\cdot The temperature at which the two phases are in equilibrium at 350bar350\:\text{bar}.

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