Answer:
168.26 [tex]Cm^3[/tex] is the volume of the cooled gas.
Explanation:
From “Charles law” for given “mass of gas” the “volume of given mass of a gas” is directly proportional to its “temperature” when expanded. [tex]\mathrm{V} \alpha T \quad \frac{V}{T}=\text { Constant }[/tex]
From “Boyle’s law” for given “mass of gas” the “volume of given mass of a gas” is inversely proportional to “pressure of given mass of a gas” when expanded. [tex]V \alpha \frac{1}{p} \quad P V=\text { Constant }[/tex]
From “Gay-Lussac law” for given “mass of gas” the “pressure of given mass of a gas” is directly proportional to “temperature of given mass of a gas” when expanded. [tex]P \alpha T \quad \frac{P}{T}=\text { Constant }[/tex]
By combining all three gas laws “Charles law”, “Boyle’s law” and “Gay-Lussac law” we know that [tex]\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}[/tex]
Given that,
[tex]\text { "Volume of gas" is } 326 \mathrm{cm}^{3}\left(\mathrm{V}_{1}\right)[/tex]
[tex]\text { "Temperature of gas" is } 22^{\circ} \mathrm{C}=22+273=295 \mathrm{K}\left(\mathrm{T}_{1}\right)[/tex]
After the gas is cooled the “Pressure of gas” is 1.90 atmosphere [tex](P_2)[/tex]
[tex]\text { "Temperature of gas" is }-10^{\circ} \mathrm{C}=-10+273=263 \mathrm{K}\left(\mathrm{T}_{2}\right)[/tex]
We need to find the “Volume of gas” after cooling.
Substitute the given values in the [tex]\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}[/tex]
[tex]\frac{1.10 \times 326}{295}=\frac{1.90 \times V_{2}}{263}[/tex]
[tex]\frac{358.6}{295}=\frac{1.90 \times V_{2}}{263}[/tex]
[tex]\frac{1.215 \times 263}{1.90}=V_{2}[/tex]
[tex]\frac{319.545}{1.90}=V_{2}[/tex]
[tex]V_{2}=168.18[/tex]
