A balloon filled with helium has a volume of 1.oo x 10^3 L at 20 degrees c. What will be the balloons volume at 30 degrees c, if the pressure surrounding the balloon remains constant?

Well, it sounds like this balloon is getting ready for a hot air balloon party! Now, let's calculate the new volume of the balloon at 30 degrees celsius while keeping the pressure constant.

To start, we can use Charles' Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature. In other words, V1/T1 = V2/T2.

Given:
V1 = 1.00 x 10^3 L (initial volume)
T1 = 20 degrees celsius (initial temperature)
T2 = 30 degrees celsius (final temperature)

Now, we can plug these values into the equation:

V1 / T1 = V2 / T2

Substituting the given values:
(1.00 x 10^3 L) / (20 degrees celsius) = V2 / (30 degrees celsius)

Cross-multiplying:
20 degrees celsius * V2 = (1.00 x 10^3 L) * (30 degrees celsius)

Simplifying:
20V2 = 30,000 L * degrees celsius

Now, let's solve for V2, which represents the final volume of the balloon at 30 degrees celsius:

V2 = 30,000 L * degrees celsius / 20 degrees celsius

V2 = 1,500 L * degrees celsius

So, the volume of the balloon at 30 degrees celsius, while keeping the pressure constant, will be 1,500 liters times the difference in temperature from 20 to 30 degrees.

To solve this problem, we can use the combined gas law equation, which relates the initial and final temperatures, as well as the initial and final volumes.

The combined gas law equation is:

P1V1 / T1 = P2V2 / T2

Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature

In this case, we are assuming that the pressure surrounding the balloon remains constant. Therefore, P1 and P2 are equal, and we can cancel them out in the equation.

Given:
V1 = 1.00 x 10^3 L (initial volume)
T1 = 20°C (initial temperature)
T2 = 30°C (final temperature)

Plugging these values into the equation, we have:

V1 / T1 = V2 / T2

Now, let's substitute the known values and solve for V2:

V1 / T1 = V2 / T2
1.00 x 10^3 L / 20°C = V2 / 30°C

To find V2, we can cross-multiply:

1.00 x 10^3 L * 30°C = V2 * 20°C

30,000 L°C = 20°C * V2

Now, we can solve for V2 by dividing both sides of the equation by 20°C:

V2 = (30,000 L°C) / 20°C

V2 = 1500 L

Therefore, the balloon's volume at 30 degrees Celsius will be 1500 L, assuming the pressure surrounding the balloon remains constant.

To solve this question, we need to use Charles's Law, which states that the volume of a gas is directly proportional to its absolute temperature (measured in Kelvins), provided that the pressure and the amount of gas remain constant.

Let's follow the steps to find the balloon's volume at 30 degrees Celsius:

Step 1: Convert the temperature from Celsius to Kelvin.
To convert from Celsius to Kelvin, we add 273 to the Celsius temperature.
For 20 degrees Celsius:
Temperature in Kelvin = 20 + 273 = 293 K

For 30 degrees Celsius:
Temperature in Kelvin = 30 + 273 = 303 K

Step 2: Use the formula of Charles's Law.
According to Charles's Law, the initial volume divided by the initial temperature is equal to the final volume divided by the final temperature.

Initial volume (V1): 1.00 x 10^3 L
Initial temperature (T1): 293 K
Final temperature (T2): 303 K
Final volume (V2): ?

V1 / T1 = V2 / T2

Step 3: Solve for V2 (the final volume).
V2 = (V1 * T2) / T1

Substituting the values:
V2 = (1.00 x 10^3 * 303) / 293

V2 ≈ 1031.24 L

So, the balloon's volume at 30 degrees Celsius, with the pressure remaining constant, will be approximately 1031.24 L.

At constant pressure, the volume of a fixed quantity of gas is proportional to the temperature, measured in °K.

To convert °C to °K, add 273.15.
So 20°C = 293.15°K, and
30°C = 303.15°K.


New volume
=1.00*10³*(303.15/293.15)
=?