57. consider two separate gas containers at the following conditions:\ncontainer a\ncontents…

57. consider two separate gas containers at the following conditions:\ncontainer a\ncontents: so₂(g)\npressure = pₐ\nmoles of gas = 1.0 mol\nvolume = 1.0 l\ntemperature = 7°c\n\ncontainer b\ncontents: unknown gas\npressure = pᵦ\nmoles of gas = 2.0 mol\nvolume = 2.0 l\ntemperature = 287°c\n\nhow is the pressure in container b related to the pressure in container a?

57. consider two separate gas containers at the following conditions:\ncontainer a\ncontents: so₂(g)\npressure = pₐ\nmoles of gas = 1.0 mol\nvolume = 1.0 l\ntemperature = 7°c\n\ncontainer b\ncontents: unknown gas\npressure = pᵦ\nmoles of gas = 2.0 mol\nvolume = 2.0 l\ntemperature = 287°c\n\nhow is the pressure in container b related to the pressure in container a?

Answer

Explanation:

Step1: Convert temperatures to Kelvin

$T_A=7 + 273=280K$, $T_B=287+ 273 = 560K$

Step2: Use the ideal - gas law $PV = nRT$

For container A: $P_A=\frac{n_ART_A}{V_A}$, substituting $n_A = 1.0mol$, $V_A=1.0L$, $T_A = 280K$ and $R$ (the ideal - gas constant). For container B: $P_B=\frac{n_BRT_B}{V_B}$, substituting $n_B = 2.0mol$, $V_B=2.0L$, $T_B = 560K$ and $R$.

Step3: Calculate the ratio $\frac{P_B}{P_A}$

$P_A=\frac{1.0mol\times R\times280K}{1.0L}=280R$ $P_B=\frac{2.0mol\times R\times560K}{2.0L}=560R$ $\frac{P_B}{P_A}=\frac{560R}{280R}=2$

Answer:

$P_B = 2P_A$