Answer
$0.964$ atm
Work Step by Step
In the gas-thermometer scale, $\displaystyle \frac{T_{2}}{T_{1}}=\frac{p_{2}}{p_{1}}\qquad(17-4).$
Figure 17.7 gives us the temperature at which $\mathrm{C}\mathrm{O}_{2}$ solidifies: $\mathrm{T}_{2}=195 \mathrm{K}$
For water, $T_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}}=\mathrm{T}_{1}=273.15 \mathrm{K}$
Apply 17-4.
$p_{2}=\displaystyle \frac{p_{1}T_{2}}{T_{1}}=\frac{(1.35 \mathrm{a}\mathrm{t}\mathrm{m} )(195\mathrm{K})}{273.15\mathrm{K}}=0.964$ atm