Chemistry help!!!

Question

The volume of a gas at 24°C and 0.39 atm

is 91 mL. What volume will the same gas

sample occupy at standard conditions?

Answer in units of mL.

Answer 1

Hello, Nereydda,

Since the moles of gas do not change in the two states, we can use a simplified expression of the gas laws:

P1V1/T1 = P2V2/T2

Rearrange to isolate the unknown, V2:

V2 = V1(T2/T1)(P1/P2)

Remeber to convert temperatures into Kelvin when using the gas laws (add 273.15 to C). Note the manner in which I grouped the terms above. They are set as ratios of the pressures and temperatures between the two states. This makes oit easier to understand what to expect. As temperature goes upo, the volume increase. As the pressure goes up, volyume decreases. In this problem, the temperature decreases around 10%, and the pressure increases by around 200%. Thus, we'd expect a decrease in volume, since the rartio of pressures has a more significant impact (P1/P2) than the decrease in temperature (T2/T1).

Standard temperature and pressure are 273.15K and 1 atm.

Plug in the numbers. We can leave the volume in units of ml. All other units cancel.

V2 = (91ml)*(297.15K/273.15K)^(0.391atm/1atm)

I get V2 = 32.7 ml.

Bob

Answer 2

Since 1 mole of an ideal gas will occupy 22.4 L at STP, we first need to find the number of moles of gas that are present. We can do this by using the ideal gas law: PV = nRT

V = volume = 91 ml x 1 L/1000 ml = 0.091 L

T = temperature = 24ºC +273 = 297K

P = pressure = 0.39 atm

R = gas constant = 0.0821 Latm/Kmol

n = moles = ?

Solving for moles, we have...

n = PV/RT = (0.39)(0.091) / (0.0821)(297) = 0.001455 moles

Converting this to volume (liters) at STP we have...

0.001455 moles x 22.4 L/mol = 0.0362 L

Converting this to mls, we have 0.0362 L x 1000 ml/L = 32.6 mls = 33 mls (2 sig. figs.)