Question

A tank contains 120.0 g chlorine gas (Cl2), which is at temperature 76.0°C and absolute pressure 5.70 ✕ 105 Pa. The temperature of the air outside the tank is 19.0°C. The molar mass of Cl2 is 70.9 g/mol. (a) What is the volume of the tank (in m3)? m3 (b) What is the internal energy of the gas (in J)? J (c) What is the work done by the gas (in J) if the temperature and pressure inside the tank drop to 31.0°C and 3.80 ✕ 105 Pa, respectively, due to a leak? (Assume that the air outside the tank can be treated as a vacuum.)

Answer 1

Answer:

Explanation:

Given that:

mass of chlorine gas = 120.0 g

the molar mass of Cl₂ = 70.9 g/mol

The number of moles of chlorine gas = mass/molar mass

The number of moles of chlorine gas = 120.0 g / 70.9 g/mol

The number of moles of chlorine gas = 1.69 mol

Pressure = $5.70 \times 10^5 \ Pa$

Temperature = 19.0°C = (273 + 76) K = 349 K

Universal Gas constant R = 8.314 J/mol/K

Using Ideal gas equation:

PV = nRT

V = nRT/P

V = (1.69 × 8.314 × 349)/ $5.70 \times 10^5$

V = 4903.68034/ $5.70 \times 10^5$

V = 0.008603 m³

Thus, the volume V of the tank =  0.008603 m³

For diatomic gas, the internal energy can be calculated by using the formula:

internal energy $E_{int}$ = $\dfrac{5}{2} \times nRT$

internal energy $E_{int}$ = $\dfrac{5}{2} \times 1.69 \times 8.314 \times 349$

internal energy $E_{int}$ = 12259.2 J

If there is a temperature and pressure drop;

the new temperature = 31.0°C = (273 + 31) K = 304 K

the new pressure = $3.80 \times 10^5 \ Pa$

We need to understand that the aftermath of the reduction in chlorine gas resulted in a decrease in temperature and pressure. Since nothing is affecting the volume of the chlorine gas, then there is no work done.

Thus, work done will be zero because there is no change in the volume or an isochoric process.