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

2 Answers

Answer:

#"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#

Mar 8, 2018

Answer:

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#