How Does Temperature and Pressure Affect Balloon Volume?

[v3ra]

A balloon is filled up with air to a volume of 1.15 L at 296.5 K.* What does the volume change to if the balloon is taken outdoors where the temperature is 278.4 K and the pressure is half of what it was indoors?

V2 = (P1V1T2) / (T1P2)


How do I find the initial pressure at 296.5 K if I know that at STP the pressure is 760 mm Hg at 273.15 K?

 

[Saitama]

A balloon is filled up with air to a volume of 1.15 L at 296.5 K.* What does the volume change to if the balloon is taken outdoors where the temperature is 278.4 K and the pressure is half of what it was indoors?

V2 = (P1V1T2) / (T1P2)


How do I find the initial pressure at 296.5 K if I know that at STP the pressure is 760 mm Hg at 273.15 K?

What remains constant in the process? Pressure, temperature, volume or moles?

If P is the initial pressure, what's the final pressure?

 

[v3ra]

The moles would remain constant... how does this help me?

 

[Saitama]

The moles would remain constant... how does this help me?

Right! So what are the initial moles and final moles? Write the expressions. Do not evaluate.

 

[DeeNos]

To find the initial pressure at 296.5 K, you can use the ideal gas law which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in Kelvin. Since we know the volume (1.15 L) and temperature (296.5 K), we can rearrange the equation to solve for pressure: P = (nRT)/V.

To find the number of moles (n), we can use the ideal gas law again, but this time with the given conditions at STP (standard temperature and pressure). At STP, the pressure is 760 mm Hg (or 1 atm) and the temperature is 273.15 K. So, we can plug these values into the ideal gas law and solve for n: n = (PV)/(RT).

Once we have the value for n, we can plug it into the first equation to solve for the initial pressure at 296.5 K. Keep in mind that the pressure is given in mm Hg, so you may need to convert it to the appropriate units (e.g. atm or kPa) depending on the units used in the first equation.