Answer
For the first method:
$$ \Delta V_1 =340\;\rm L $$
$$ \Delta V_2 =340\;\rm L $$
For the second method:
$$ \Delta V_1 =578\;\rm L $$
$$ \Delta V_2 =366\;\rm L $$
Work Step by Step
$\Rightarrow$ The first method for calculating the thermal expansion under constant pressure of 1 atm is given by
$$\Delta V =V_0\beta \Delta T$$
whereas $\beta$ is the thermal coefficient of expansion (see table 13-1).
$\rightarrow$ In the first case, when raising the temperature of an ideal gas from $-100^\circ \;\rm C$ to $0^\circ \;\rm C$
$$\Delta V_1 =1000\cdot 3400\times10^{-6}\left(0-[-100]\right)
$$
$$\boxed{\Delta V_1 =340\;\rm L}$$
$\rightarrow$ In the second case, when raising the temperature of an ideal gas from $0^\circ \;\rm C$ to $100^\circ \;\rm C$
$$\Delta V_2 =1000\cdot 3400\times10^{-6}\left(100-0\right)
$$
$$\boxed{\Delta V_2 =340\;\rm L}$$
We know that this method will be accurate if the change in volume is small in comparison to the initial volume (which is not suitable for gases at all) which does not occur here.
Thus, these results are not accurate.
$\Rightarrow$ The second method for calculating the thermal expansion of an ideal gas under the constant pressure of 1 atm is given by the ideal gas law.
$$PV=nRT$$
To compare the gas after heating for cooling it;
$$\dfrac{P_iV_i}{P_fV_f}=\dfrac{nRT_i}{nRT_f }$$
Thus, under the same pressure; $P_i=P_f$
$$\dfrac{ V_i}{ V_f}=\dfrac{ T_i}{ T_f }$$
Solving for $V_f$;
$$V_f=V_i\;\dfrac{ T_f}{ T_i } $$
Hence,
$$\Delta V=V_f-V_i$$
Plugging $V_f$ from the previous formula;
$$\Delta V=V_i\;\dfrac{ T_f}{ T_i }-V_i=V_i\left(\dfrac{ T_f}{ T_i }-1\right)$$
Thus,
$$\Delta V= V_i\left(\dfrac{ T_f}{ T_i }-1\right)$$
$\rightarrow$ In the first case, when raising the temperature of an ideal gas from $173\;\rm C$ to $273 \;\rm C$
We converted temperature to Kelvin since we are not using the change in temperature in this case.
$$\Delta V_1= 1000\left(\dfrac{ 273}{ 173}-1\right)$$
$$\boxed{\Delta V_1 =578\;\rm L}$$
$\rightarrow$ In the first case, when raising the temperature of an ideal gas from $273\;\rm C$ to $373 \;\rm C$
$$\Delta V_2= 1000\left(\dfrac{ 373}{ 273}-1\right)$$
$$\boxed{\Delta V_2 =366\;\rm L}$$
The two methods differ since the first method considers the relation between volume and temperature is linear while the second method considers the relation between volume and temperature is not linear.
