Question

A 1.7 L coffee thermos is at a pressure of 1.2 atm and a temperature of 20ºC. Hot coffee is added to the thermos and it reduces the volume of gas to 1.1 L and increases the temperature to 90ºC. What is the new pressure in the coffee thermos?

Answer 1

1) List the known and unkown values

Initial conditions

Volume: 1.7 L

Pressure: 1.2 atm

Temperature: 20ºC

Final conditions

Volume: 1.1 L

Temperature: 90ºC

Pressure:

2) Set the equation

`(P1V1)/(T1)=(P2V2)/(T2)`

3) Convert Celsius to Kelvin

`K=ºC+273.15`

Initial temperature

`K=20ºC+273.15=293.15\text{ }K`

Final temperature

`K=90ºC+273.15=363.15\text{ }K`

4) Plug in the known values

`\frac{(1.2\text{ }atm)*(1.7\text{ }L)}{293.15\text{ }K}=\frac{(P2)*(1.1\text{ }L)}{363.15\text{ }K}`

.

`(\frac{(1.2\text{ }atm)*(1.7\text{ }L)}{293.15\text{ }K})*363.15\text{ }K=(\frac{(P2)*(1.1\text{ }L)}{363.15\text{ }K})*363.15\text{ }K`
`(\frac{(1.2\text{ }atm)*(1.7\text{ }L)}{293.15\text{ }K})*363.15\text{ }K=(P2)*(1.1\text{ }L)`

.

`(\frac{(1.2\text{ }atm)*(1.7\text{ }L)*(363.15\text{ }K)}{293.15\text{ }K})*\frac{1}{1.1\text{ }L}=\frac{(P2)*(1.1\text{ }L)}{1.1\text{ }L}`

.

`P2=\frac{(1.2\text{ }atm)*(1.7\text{ }L)*(363.15\text{ }K)}{(293.15\text{ }K)*(1.1\text{ }L)}=2.297\text{ }atm`

The new pressure in the coffee thermos is 2.3 atm.

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