Question

a sample of m = 0.125 kg of xenon (mono - atomic gas) is contained in a rigid metal cylinder, big enough that the xenon can be modeled as an ideal gas, at a temperature of t = 25 c. the cylinder is moved outside on a hot summer day. as the xenon comes to equilibrium by reaching the temperature of its surroundings at t (unknown), a q = 185 j of heat are conducted to it through the cylinder walls. what is the equilibrium temperature? consider the process is isovolumetric or isochoric.
0v = 0, w = \\(\int pdv = 0\\)
\\(q=\delta e+w\\) \\(\
ightarrow\\) \\(q = \delta e=185 j\\).
\\(\delta e=\frac{3}{2}nr\delta t = q\\)
\\(\delta t=\frac{2q}{3nr}\\)
\\(t_2 - 298=\frac{2q}{3nr}\\)
\\(t_2=\frac{2q}{3nr}+298=\frac{2\times185}{3\times?\times8.31}+298 =?\\)
\\(t_1=(25 + 273)k\\) \\(t_2 =?\\)
\\(p_1\\) \\(p_2\\)
\\(\delta t=t_2 - t_1=t_2 - 298\\)
\\(n = \frac{m}{m}=\frac{0.125}{m(xe)}\\) \\(n =?\\)

Explanation:

Step1: Calculate the number of moles of xenon

The molar mass of xenon ($M(Xe)$) is approximately $131.29\ g/mol = 0.13129\ kg/mol$. The number of moles $n=\frac{m}{M}$, where $m = 0.125\ kg$. So $n=\frac{0.125}{0.13129}\ mol$.

Step2: Recall the molar - specific heat for a mono - atomic ideal gas

For a mono - atomic ideal gas at constant volume, the molar - specific heat $C_{V}=\frac{3}{2}R$, and the change in internal energy $\Delta E = nC_{V}\Delta T$. Since $\Delta E=Q$ (because $W = 0$ for an isochoric process), and $C_{V}=\frac{3}{2}R$, we have $Q = n\times\frac{3}{2}R\times\Delta T$. Then $\Delta T=\frac{2Q}{3nR}$.

Step3: Calculate the final temperature

We know that $T_1=(25 + 273)K=298K$, and $\Delta T=T_2 - T_1$. So $T_2=T_1+\frac{2Q}{3nR}$. Substitute $Q = 185J$, $n=\frac{0.125}{0.13129}\ mol$ and $R = 8.31\ J/(mol\cdot K)$ into the formula.
First, calculate $n=\frac{0.125}{0.13129}\approx0.952\ mol$.
Then $\Delta T=\frac{2\times185}{3\times0.952\times8.31}=\frac{370}{23.73}\approx15.6\ K$.
$T_2=298 + 15.6=313.6\ K$.

Answer:

$313.6\ K$