Question

A) Compute the specific heat capacity at constant volume of nitrogen (N2) gas. The molar mass of N2 is 28.0 g/mol.

the answer is: 741 J/(kg*k)

B) You warm 1.55kg of water at a constant volume from 23.0?C to 29.0?C in a kettle. For the same amount of heat, how many kilograms of 23.0?C air would you be able to warm to 29.0?C? Make the simplifying assumption that air is 100% N2.

C) What volume would this air occupy at 23.0?C and a pressure of 1.05atm ?

Answer 1

Solution:

(A) The constant volume of nitrogen gas is, \(C_{V}=\frac{5}{2} R\)

\(=\frac{5}{2}(8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})\)

\(=20.79 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\)

The specific heat capacity of the constant volume of nitrogen gas is. \(\begin{aligned} c_{N_{2}} &=\frac{C_{N_{2}}}{M} \\ &=\frac{20.79 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}{0.028 \mathrm{~kg} / \mathrm{mol}} \\ &=742 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \end{aligned}\)

Hence, the specific heat capacity of the constant volume of nitrogen gas is \(742 \mathrm{~J} / \mathrm{kg}\).

(B) heat energy of the required to warm the water is, \(Q=m c_{w} \Delta T\)

\(=(1.55 \mathrm{~kg})(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(302 \mathrm{~K}-296 \mathrm{~K})\)

\(=38911.2 \mathrm{~J}\)

\(=3.89 \times 10^{4} \mathrm{~J}\)

The mass of the warm of air is, \(\begin{aligned} m_{a r} &=\frac{Q}{C_{N_{2}} \Delta T} \\ &=\frac{3.86 \times 10^{4} \mathrm{~J}}{742 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \times 6 \mathrm{~K}} \\ &=8.67 \mathrm{~kg} \end{aligned}\)

Hence, the mass of the warm of air is \(8.67 \mathrm{~kg}\)

(C) From the ideal

The number of moles is, \(n=\frac{m}{M}\)

\(\begin{aligned} P V &=\frac{m R T}{M} \\ V &=\frac{m R T}{M P} \end{aligned}\)

\(=7.16 \mathrm{~m}^{3}\)

Hence, the volume of the air is \(7.16 \mathrm{~m}^{3}\).