Question

1. 1.38 mol sample of carbon dioxide gas at a temperature of 30.0 °C is found to occupy a volume of 22.6 liters. The pressure of this

gas sample is _____
mm Hg.

2. A 5.32 gram sample of carbon dioxide gas has a volume of 826 milliliters at a pressure of 2.11 atm. The temperature of the CO2 gas sample is ____ °C.

Answer 1

Answer: 1. the pressure of the carbon dioxide gas sample is approximately 2.00 atm. 2.the temperature of the carbon dioxide gas sample is approximately 60.7 °C.

Step-by-step explanation:

1. To find the pressure of the carbon dioxide gas sample, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, let's convert the temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15

T(K) = 30.0 °C + 273.15

T(K) = 303.15 K

Now, let's substitute the values into the equation:

P * 22.6 L = 1.38 mol * 0.0821 L·atm/(mol·K) * 303.15 K

Simplifying the equation:

P = (1.38 mol * 0.0821 L·atm/(mol·K) * 303.15 K) / 22.6 L

Calculating the pressure:

P ≈ 2.00 atm

2. To find the temperature of the carbon dioxide gas sample, we can rearrange the ideal gas law equation as follows:

T = (PV) / (nR)

First, let's convert the volume from milliliters to liters:

V(L) = V(mL) / 1000

V(L) = 826 mL / 1000

V(L) ≈ 0.826 L

Now, let's substitute the values into the equation:

T = (2.11 atm * 0.826 L) / (5.32 g / (44.01 g/mol) * 0.0821 L·atm/(mol·K))

Simplifying the equation:

T = (2.11 atm * 0.826 L) / (0.0832 mol * 0.0821 L·atm/(mol·K))

Calculating the temperature:

T ≈ 60.7 °C