Question

A sample of nitrogen gas at a pressure of 3 atm inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. What is the pressure of the nitrogen after its temperature is increased? Answer atm

Answer 1

`"3.31 atm"`

Explanation:

Gay-Lussac's Law states `P_1/T_1 = P_2/T_2`

So let's take the `"N"_"2(g)"` and calculate the `P_1` divided by `T_1`, which is `"3 atm"` divided by `"293 K"`. Then we take `P_2` divided by `T_2` which is `P_2` divided by `"323 Kelvin"`.

`"3 atm"/(293 cancel"Kelvin") = (P_2)/(323 cancel"Kelvin")`

`"0.010239 atm" = (P_2)/323`

Then cross multiply, imagine the denominator of `0.010239` is `1`, since `0.010239` divided by `1` is `0.010239`.

`"0.010239 atm" xx 323 = P_2 xx 1`

`"3.31 atm" = P_2`

Answer 2

See Below

Explanation:

Rigid means the volume doesn't change. As long as it is sealed, the moles don't change, either.

`n_1 = n_2......n="PV"/"RT"`

`(P_1V_1)/(RT_1) = (P_2V_2)/(RT_2)`

R and R are the same, so they cancel. V1 and V2 are the same (rigid container), so they cancel. You are left with

`(P_1)/(T_1) = (P_2)/(T_2)`

Change 20C to 293K (T1) and 50C to 323K (T2) and solve for P2

`P_2 = (P_1)/(T_1)xxT_2`
`P_2 = (3atm)/(293)xx323`

`P_2 = 3.31 atm`