Question

1. A sample of gas is trapped in a rigid container at 1.15 atm and 31 oC. If the container is heated until the pressure in the container is 2.56 atm, what is the temperature of the gas in the container (in oC)?

2.A balloon containing helium gas has a volume of 3.15 L at room temperature 23 oC. The balloon is heated to a temperature of 82 oC. Assuming no change in pressure, what is the new volume of the balloon?
3. A 26.8 L tank of nitrogen gas is at 26.0 oC and 1.6 atm. If the temperature stays at 26.0 oC and the volume is compressed to 12.2 L, what is the new pressure?
4. An expandable balloon containing 1.28 moles of helium gas and has a volume of 1.595 L. A certain amount of argon gas is added to the balloon and it expands to 5.829 L. If there is no change in the temperature and pressure, how many moles of argon gas were added to the balloon?
Swamped with LSAT studies ,a ll help is appreciated ! =)

Answer 1

The four questions evaluate the application of gas laws (Gay-Lussac's, Charles's, Boyle's, and Avogadro's). Their solutions result in approximate values of 400.55°C for the increased temperature under pressure, 3.52 L for the new volume after heating the balloon, 3.51 atm for the increased pressure after compressing the gas, and addition of approximately 3.39 moles of argon.

Step-by-step explanation:

To answer these questions, we need to use the gas laws, which describe how gases behave under varying conditions.

For question 1, we use Gay-Lussac's law, which states that the pressure of a gas is directly proportional to its temperature if the volume remains constant. We can use the equation P1/T1 = P2/T2, but first we need to convert the temperature to Kelvin: T1 = 31°C + 273.15 = 304.15 K. Then, solving for T2 (the final temperature), we get T2 = P2*T1 / P1 = 2.56 atm * 304.15 K / 1.15 atm = 673.7 K, which when converted back to Celsius, equals 400.55°C.

For question 2, we use Charles's law (V1/T1 = V2/T2). Again, our temperatures must be in Kelvin. Thus the calculation is: V2 = V1 * T2 / T1 = 3.15L * (82°C + 273.15) / (23°C + 273.15) = 3.52 L.

For question 3, we apply Boyle's law (P1*V1 = P2*V2) which states that at a constant temperature, the product of the volume and pressure of a given amount of gas remains constant. Thus we calculate: P2 = P1*V1 / V2 = 1.6 atm * 26.8 L / 12.2 L = 3.51 atm.

Lastly, for question 4, since the temperature and pressure remained constant, we can use Avogadro's law (n1/V1 = n2/V2) which states that at a fixed temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas. Solving for n2 (our final moles of gas), we have: n2 = n1 * V2 / V1 = 1.28 moles * 5.829 L / 1.595 L = 4.67 moles. But since we originally had 1.28 moles of Helium, the amount of Argon added is 4.67 moles - 1.28 moles = 3.39 moles.

Learn more about Gas Laws