Question

I really need help with question one and two. I don’t remember how to do it and my notes aren’t making matters better

Answer 1

#1. Answer:

0.449 mole

Solution and explanation:

We are given;

Volume of oxygen gas,V = 75.0 mL or 0.075 L

Pressure, P = 145 atm

Temperature, T = 22° C or 22 + 273 = 295 K

We are required to calculate the moles of Oxygen gas

We are going to use the ideal gas equation which states;

PV =nRT where, P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant,(0.082057 L atm/mol.K) and T is the temperature.

Therefore, to get n we use;

$n=\frac{PV}{RT}$

$n=\frac{(145)(0.075)}{(0.082057)(295)}$

$n=0.449 moles$

Therefore; the number of moles of Oxygen gas is 0.449 mole

# 2. Answer

35,410.99 torr

Solution and Explanation:

We are given

Initial Pressure of the gas, P1 = 30,400 torr

Initial temperature, T1 = 0°C or 273 K

Final temperature, T2 = 45°C or 318 K

We are required to find the final pressure, P1,

• According to the pressure Law, pressure of a gas and its absolute temperature vary direct proportionally to each other at constant volume.

Therefore,

$P\alpha T$

At varying temperature and pressure;

$\frac{P1}{T1} =\frac{P2}{T2}$

Thus;

$P2 = T2(\frac{P1}{T1})$

$P2= 318(\frac{30,400}{273})$

$= 35,410.99 torr$

Therefore; the new pressure in the container is 35,410.99 torr