- A balloon containing 5.00 L of argon has a pressure of 102,390 Pa and a temperature of 27.°C. The balloon is pulled under water where the pressure of the balloon becomes 405,300 Pa and the volume drops to 1.17 L. What is the Celsius temperature under these conditions? Assume the moles remain constant.
Ideal and Real Gases
Ideal gases obey conditions of the general gas laws under all states of pressure and temperature. Ideal gases are also named perfect gases. The attributes of ideal gases are as follows,
Gas Laws
Gas laws describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. The very first observations about the physical properties of gases was made by Robert Boyle in 1662. Later discoveries were made by Charles, Gay-Lussac, Avogadro, and others. Eventually, these observations were combined to produce the ideal gas law.
Gaseous State
It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed forms of matter studied are solids, gases and liquids. The best example of a substance that is present in different states is water. It is solid ice, gaseous vapor or steam and liquid water depending on the temperature and pressure conditions. This is due to the difference in the intermolecular forces and distances. The occurrence of three different phases is due to the difference in the two major forces, the force which tends to tightly hold molecules i.e., forces of attraction and the disruptive forces obtained from the thermal energy of molecules.
![**Problem 10: Gas Laws in Various Conditions**
**Context:**
This problem involves a scenario where a balloon containing a gas (argon) is subjected to changes in pressure, volume, and temperature. It requires the application of the combined gas law to find the unknown temperature under new conditions.
**Given Data:**
- Initial volume of argon (V1): 5.00 L
- Initial pressure of argon (P1): 102,390 Pa
- Initial temperature of argon (T1): 27°C (300.15 K)
- New pressure of the balloon (P2): 405,300 Pa
- New volume of the balloon (V2): 1.17 L
**Objective:**
Determine the new Celsius temperature (T2) when the pressure of the balloon becomes 405,300 Pa and the volume drops to 1.17 L, assuming the moles of argon remain constant.
**Solution Process:**
To find the new temperature, we use the combined gas law:
\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \]
Rearrange to solve for \( T_2 \):
\[ T_2 = \frac{P_2 \times V_2 \times T_1}{P_1 \times V_1} \]
**Calculation:**
Convert initial temperature to Kelvin:
\[ T_1 = 27°C + 273.15 = 300.15 \, K \]
Substitute the given values into the equation:
\[ T_2 = \frac{405,300 \, \text{Pa} \times 1.17 \, \text{L} \times 300.15 \, \text{K}}{102,390 \, \text{Pa} \times 5.00 \, \text{L}} \]
\[ T_2 = \frac{405,300 \times 1.17 \times 300.15}{102,390 \times 5.00} \]
\[ T_2 \approx \frac{142,349,365.5}{511,950} \]
\[ T_2 \approx 278.0 \, K \]
Convert \( T_2 \) back to Celsius:
\[ T_2 = 278.0 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec9904e6-dc28-47e5-ba82-a0b2892d5fea%2F3b981918-56c6-464d-85ca-31278c8285be%2Frt00ag_processed.png&w=3840&q=75)
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