Answer to Question #210076 in Inorganic Chemistry for DIANA

Question

Calculate each of the following quantities for an ideal gas: a) The volume of the gas in litre, if 2.57 moles have a pressure of 1.86 atm at a temperature of -12ºC. b) The absolute temperature of a gas at which 6.79 X 10 2 moles occupies 164 ml at 893 torr. c) The pressure in atm , if 8.25 X 10 2 moles occupies 255 mL at 15°C. d) If oxygen gas is used, calculate the mass of the oxygen gas in grams, if 5.49 L at 45ºC has a pressure of 11.25 kPa.

Answer 1

a) No of moles(n) = 2.57 mol

Pressure of gas(P) = 1.86 atm

Temperature(T) = - 12C = 261K

Gas Constant(R) = 0.0821 L.atm/mol.K

By Ideal Gas Law, PV = nRT

So, V = nRT/P =

2.57 × 0.0821 × 261 / 1.86 = 29.607 L

So, The volume of gas = 29.607 L .

b) Volume of gas (V) = 164mL = 0.164 L

No of moles (n) = 6.79 × 10-2 mol = 0.0679 mol

Pressure of gas (P) = 893 torr

Gas Constant(R)= 62.36 L.Torr/mol.K

By Ideal Gas law, PV = nRT

So, T = PV/nR =

893 × 0.164 / 0.0679 × 62.36

So, the Temperature of gas(T) = 34.587 K

c) No of moles (n) =

8.25 × 10-2 mol = 0.0825 mol

Volume of gas (V) = 255mL = 0.255L

Temperature (T) = 15C = 288 K

Gas Constant(R) = 0.0821L.atm/mol.K

By ideal gas Law, PV = nRT

So, P = nRT/V =

0.0825 × 0.0821 × 288 / 0.255

So, the pressure of gas = 7.649 atm

d) Volume of Oxygen gas (V) = 5.49L

Temperature(T) = 45C = 318 K

Pressure of gas (P)= 11.25 kPa

Gas Constant (R) = 8.314 L.kPa/mol.K

By Ideal gas law, PV=nRT

So, n= PV/RT =

11.25 × 5.49/ 8.314 × 318

So, n = 0.023 mol

Molar mass of Oxygen gas = 32 g/mol

The mass of oxygen gas = 0.023 × 32 = 0.736 g .