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}}{p_{1}}\\) 3 putting in the numbers t₂ = \\(\frac{__5__ k __6__ atm __7__ l}{__9__ atm __10__ l}\\) = 8 k a. n₂ b. 20.0 c. 473.15 d. 323.15 e. 1.000 f. p₁ g. p₂ h. v₁ i. v₂ j. n₁ k. t₁ l. t₂ m. 1.464

Explanation:

Step1: Convert Celsius to Kelvin

$T_1 = 25.00 + 273.15=298.15$ K

Step2: Use the combined gas law

Since $n_1 = n_2$, the combined gas law $\frac{P_1V_1}{n_1T_1}=\frac{P_2V_2}{n_2T_2}$ simplifies to $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$.
Solving for $T_2$, we get $T_2=\frac{P_2V_2T_1}{P_1V_1}$.

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.
$T_2=\frac{2.000\times1.000\times298.15}{1.000\times6.000}$
$T_2=\frac{596.3}{6}=99.383\approx99.4$ K (This part seems there might be a mismatch with the given options, but following the formula structure in the problem):
If we follow the formula structure in the problem:
$T_2=\frac{T_1P_2V_2}{P_1V_1}$, substituting $T_1 = 298.15$ (which is $25 + 273.15$), $P_1 = 1.000$, $V_1 = 6.000$, $P_2 = 2.000$, $V_2 = 1.000$
$T_2=\frac{298.15\times2.000\times1.000}{1.000\times6.000}=\frac{596.3}{6}=99.38\approx99.4$

For the blanks:

1. $T_1=298.15$ (from $25 + 273.15$)
2. $n_1=n_2$
3. $V_2$
4. $V_1$
5. $T_1 = 298.15$
6. $P_2=2.000$
7. $V_2 = 1.000$
8. (calculated as above)
9. $P_1 = 1.000$
10. $V_1=6.000$

Answer:

1. $298.15$; 2. $n_2$; 3. $V_2$; 4. $V_1$; 5. $298.15$; 6. $2.000$; 7. $1.000$; 8. $\frac{298.15\times2.000\times1.000}{1.000\times6.000}$; 9. $1.000$; 10. $6.000$