Question

The volume of gas at 1.10 atm was measured at 22°C and 326 cm³. What will be the volume of the gas if cooled to -10°C and the pressure is adjusted to 1.90 atm?

Answer 1

168.26
`Cm^3` is the volume of the cooled gas.

Step-by-step 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.
`\mathrm{V} \alpha T \quad (V)/(T)=\text { Constant }`

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.
`V \alpha (1)/(p) \quad P V=\text { Constant }`

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.
`P \alpha T \quad (P)/(T)=\text { Constant }`

By combining all three gas laws “Charles law”, “Boyle’s law” and “Gay-Lussac law” we know that
`(P_(1) V_(1))/(T_(1))=(P_(2) V_(2))/(T_(2))`

Given that,

`\text {`

`\text {`

After the gas is cooled the “Pressure of gas” is 1.90 atmosphere
`(P_2)`

`\text {`

We need to find the “Volume of gas” after cooling.

Substitute the given values in the
`(P_(1) V_(1))/(T_(1))=(P_(2) V_(2))/(T_(2))`

`(1.10 * 326)/(295)=(1.90 * V_(2))/(263)`

`(358.6)/(295)=(1.90 * V_(2))/(263)`

`(1.215 * 263)/(1.90)=V_(2)`

`(319.545)/(1.90)=V_(2)`

`V_(2)=168.18`