# 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

Mar 8, 2018

$\text{3.31 atm}$

#### Explanation:

Gay-Lussac's Law states ${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

So let's take the $\text{N"_"2(g)}$ and calculate the ${P}_{1}$ divided by ${T}_{1}$, which is $\text{3 atm}$ divided by $\text{293 K}$. Then we take ${P}_{2}$ divided by ${T}_{2}$ which is ${P}_{2}$ divided by $\text{323 Kelvin}$.

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

$\text{0.010239 atm} = \frac{{P}_{2}}{323}$

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

$\text{0.010239 atm} \times 323 = {P}_{2} \times 1$

$\text{3.31 atm} = {P}_{2}$

Mar 8, 2018

See Below

#### Explanation:

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

${n}_{1} = {n}_{2.} \ldots . . n = \text{PV"/"RT}$

$\frac{{P}_{1} {V}_{1}}{R {T}_{1}} = \frac{{P}_{2} {V}_{2}}{R {T}_{2}}$

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

$\frac{{P}_{1}}{{T}_{1}} = \frac{{P}_{2}}{{T}_{2}}$

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

${P}_{2} = \frac{{P}_{1}}{{T}_{1}} \times {T}_{2}$
${P}_{2} = \frac{3 a t m}{293} \times 323$

${P}_{2} = 3.31 a t m$