Question

give the temperature when 1.000 atm of ar in 6.000 l at 25.00 c is compressed to 1.000 l and 2.000 atm. this is a change in conditions problem, we need to define the variables: v₁ = 6.000 l t₁ = 1 k p₁ = 1.000 atm v₂ = 1.000 l t₂ = ? k p₂ = 2.000 atm the problem doesnt mention the moles, n, thus the amount of moles was not changed and n₁ = 2 the equation simplifies to \\(\frac{p_{2}v_{2}}{n_{2}t_{2}}=\frac{p_{1}v_{1}}{n_{1}t_{1}}\\) \\(\frac{p_{2}v_{2}}{t_{2}}=\frac{p_{1}v_{1}}{t_{1}}\\) solving for t₂ = \\(\frac{t_{1}p_{2}v_{2}}{p_{1}v_{1}}\\) putting in the numbers t₂ = \\(\frac{298.15 k \times 2.000 atm \times 1.000 l}{1.000 atm \times 6.000 l}\\) = 99.38 k

Explanation:

Step1: Convert \(T_1\) from Celsius to Kelvin

\(T_1 = 25.00+273.15=298.15\) K

Step2: Use the ideal gas law \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\) (since \(n_1 = n_2\))

\(T_2=\frac{P_2V_2T_1}{P_1V_1}\)

Step3: Substitute the values \(P_1 = 1.000\) atm, \(V_1 = 6.000\) L, \(T_1 = 298.15\) K, \(P_2 = 2.000\) atm, \(V_2 = 1.000\) L

\(T_2=\frac{2.000\times1.000\times298.15}{1.000\times6.000}\)

Step4: Calculate the value of \(T_2\)

\(T_2=\frac{2\times298.15}{6}\approx99.38\) K

Answer:

\(T_2 = 99.38\) K