Answer
See the detailed answer below.
Work Step by Step
a) First of all, we need to find the number of moles of the 10 g-dry ice $ (\rm CO_2)$ sample where we know that the number of moles is given by
$$n=\dfrac{m}{M}$$
where $m$ is the mass of the sample and $M$ is the atomic mass of one molecule.
Thus,
$$n=\dfrac{m}{M_C+2M_O}$$
Plugging the known;
$$n=\dfrac{10}{12+2(16)}=\bf0.2273 \;\rm mol$$
We know, for an ideal gas, that
$$PV=nRT$$
Hence, the gas pressure at 0$^\circ $C is
$$P_1=\dfrac{nRT_1}{V_1} $$
Plugging the known;
$$P_1=\dfrac{(0.2273)(8.31)(0+273)}{10,000\times 10^{-6}} =\bf 5.16\times 10^4\;\rm Pa$$
$$P_1=0.509\;\rm atm\approx \color{red}{\bf 0.51}\;\rm atm$$
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b)
In the second isothermal compression process, the temperature is constant. So
$$P_1V_1=P_2 V_2$$
Thus, the volume at the end of this stage is given by
$$V_2=\dfrac{P_1V_1}{P_2 }$$
Plugging the known;
$$V_2=\dfrac{(0.509)(10,000)}{3}=\bf 1.697\times 10^3\;\rm cm^3$$
In the third isobaric process where the pressure constant is constant.
So
$$\dfrac{V_2}{T_2}=\dfrac{V_3}{T_3}$$
where $T_2=T_1$ since it was an isothermal process.
$$\dfrac{V_2}{T_1}=\dfrac{V_3}{T_3}$$
Hence,
$$T_2=\dfrac{V_3T_1}{V_2}$$
And to find it in Celsius,
$$T_2=\dfrac{V_3T_1}{V_2}-273$$
Plugging the known
$$T_2=\dfrac{(1000)(0+273)}{( 1.697\times 10^3)}-273$$
$$T_2=\color{red}{\bf-112}^\circ\rm C$$
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c) We have here 2 stages, an isothermal stage followed by an isobaric stage.
$\bullet$ The isothermal stage starts from $\rm(10,000 \;cm^3,0.509\; atm)$ and ends at $\rm( 1697\;cm^3,3\; atm)$
$\bullet$ The isobaric stage starts from $\rm( 1697\;cm^3,3\; atm)$ and ends at $\rm(1000\;cm^3,3\; atm )$
See the graph below.
