chemistry that's eluding me

Question

If a 3.6 moles of an ideal gas occupies 150 mL tank at a temperature of -5 °C, what is the pressure inside the tank?

Answer 1

PV = nRT

P = pressure = ?

V = Volume = 150 ml x 1 L/1000 ml = 0.150 L

n = moles = 3.6

R = gas constant = 0.0821 Latm/Kmol

T = temperature = -5ºC + 273 = 268K

Solving for P, we have

P = nRT/V = (3.6 moles)(0.0821 Latm/Kmol)(268K)/0.150 L

P = 528 atm

Answer 2

One mole of an ideal gas will occupy a volume of 22.4 liters at STP (Standard Temperature and Pressure, 0°C and one atmosphere pressure).

We can use the ideal gas law....we have an ideal gas PV=nRT

Convert 0 C to 273.15 K and -5 C to 273.15 - 5 the degree Kelvin and Celsius divisions are the same magnitude.

P is directly proportional to temperature and number of molecules of gas and inversely proportional to volume.

Lets say we kept pressure and temperature at stp and just changed things to 3.6 moles of gas

We have 3.6 x 22.4 L of gas at STP or 80.6 L of gas or to sig figs 81 L

Lets say we drop the temperature by 5 C we have 80.6 L scaled by the temperature ratio 268.15/273.15 at 1 atm pressure or 79 L of gas.

Now we want to stuff that into our 150 mL tank so our Pressure scales by 79 L/0.150 L and is 530 atm

Does this seem sensible...lets look at the pressure involved in just stuffing 22.4 L of gas at STP into our vessel and forget changing the temperature... 22.4/0.150 = 149 ....150 x 3 is 450....so the pressure we calculated is quite reasonable.