Question

0.680 moles of O2(g) are added to 4.00 L flask and the internal pressure is measured at 14.00 atm. What is the temperature of the gas under these conditions (in °C)?___ °C

Answer 1

Temperature = 729.93 °C

Step-by-step explanation

Given

Number of moles = 0.680 mol

Volume = 4.00 L

Pressure = 14.00 atm

We know the gas constant = 0.0821 L.atm/k.mol

Required: Temperature

Solution

To solve this problem we will use the ideal gas law

PV = nRT

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

We want to solve for T:

T = PV/nR

T = (14.00 atm x 4.00 L)/(0.680 mol x 0.0821 L.atm/k.mol)

T = 1003.1 K

We want the temperature in °C

0°C + 273.15 = 273,15K

Therefore

°C = 1003.1 K - 273.15 = 729.93 °C