Question

calculate the volume at stp for 4.000 l balloon filled with oxygen at 2.000 atm and 25.00 °c. this is a change in conditions problem, we need to define the variables: v₁ = 4.000 l t₁ = 1 k p₁ = 2.000 atm v₂ =? l t₂ = 2 k p₂ = 1.000 atm the problem doesnt mention the moles, n, thus the amount of moles was not changed and n₁ = 3. 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 v₂ = \\(\frac{p_{1}v_{1}t_{2}}{t_{1}p_{2}}\\). putting in the numbers v₂ = \\(\frac{2.000 atm\times4.000 l\times273.15 k}{298.15 k\times1.000 atm}=7.329 l\\)

Explanation:

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

To convert \(T_1 = 25.00^{\circ}C\) to Kelvin, use the formula \(T(K)=T(^{\circ}C)+273.15\). So \(T_1=25.00 + 273.15=298.15\) K.

Step2: Identify \(T_2\) (STP temperature)

At STP (Standard Temperature and Pressure), the temperature \(T_2 = 273.15\) K.

Step3: Determine \(n_1\) and \(n_2\)

Since the amount of gas (moles) is not changed, \(n_1=n_2\).

Step4: Simplify the ideal - gas law equation

From \(\frac{P_1V_1}{n_1T_1}=\frac{P_2V_2}{n_2T_2}\), and \(n_1 = n_2\), the equation simplifies to \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), solving for \(V_2=\frac{P_1V_1T_2}{P_2T_1}\).

Step5: Substitute the values

Substitute \(P_1 = 2.000\) atm, \(V_1 = 4.000\) L, \(T_1=298.15\) K, \(P_2 = 1.000\) atm, \(T_2 = 273.15\) K into \(V_2=\frac{P_1V_1T_2}{P_2T_1}\).

Answer:

1. \(298.15\)
2. \(273.15\)
3. \(n_2\)
4. \(P_2T_2\)
5. \(P_1 = 2.000\) atm, \(V_1 = 4.000\) L, \(T_1 = 298.15\) K, \(P_2 = 1.000\) atm, \(T_2 = 273.15\) K
6. \(2.000\)
7. \(4.000\)
8. \(273.15\)
9. \(298.15\)
10. \(1.000\)