Question

a 5.50 L air sample is at STP inside of a metal cylinder. what is the new temperature in Celsius if the pressure is lowered to 99.5 kPa?

Answer 1

Step-by-step explanation:

To solve this problem, we can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

At STP (standard temperature and pressure), the pressure is 101.325 kPa and the temperature is 273.15 K.

We can use this information to find the initial number of moles of air in the cylinder:

PV = nRT

(101.325 kPa) (5.50 L) = n (8.314 J/mol·K) (273.15 K)

n = 0.226 mol

Now, we can use the same equation to find the new temperature, assuming the number of moles of air remains constant and the volume of the cylinder is unchanged:

PV = nRT

(99.5 kPa) (5.50 L) = (0.226 mol) (8.314 J/mol·K) T

T = (99.5 kPa) (5.50 L) / (0.226 mol) (8.314 J/mol·K)

T = 301.8 K

Therefore, the new temperature of the air sample in Celsius is:

T = 301.8 K - 273.15 K = 28.65°C

So the new temperature is 28.65°C.