Question

g compute the specific heat capacity at constant volume of nitrogen gas. the molar mass of N2 is 29.0 You warm 1.8 kg ov water at a constant volume from 21 C to 30.5 C in a kettle

Answer 1

Complete question:

(a) compute the specific heat capacity at constant volume of nitrogen gas. the molar mass of N₂ is 29.0 You warm 1.8 kg of water at a constant volume 1.00 L from  21 C to 30.5 C in a kettle. For the same amount of heat, how many kilograms of 21∘C air would you be able to warm to 30.5∘C ?

(b) What volume (in liters) would this air occupy at 21∘C and a pressure of 1.00 atm? Make the simplifying assumption that air is 100% N₂​

Answer:

(a) The specific heat capacity of N₂ is 715.86 J/kg.K

(b) The volume the air occupy at 21∘C is 8784.29 Liters

Explanation:

Given;

M is the molar mass of N₂ = 29 x 10⁻³ kg/mol

specific heat of N₂ at constant volume, Cv = 20.76 J/mol.K

(a)

The specific heat capacity of N₂ is calculated as;

$C = \frac{C_v}{M} \\\\C = \frac{20.76}{29 *10^{-3}} \\\\C = 715.86 \ J/kg.K$

(b) heat capacity of water;

Q = mcΔθ

where;

c is the specific heat capacity of water = 4200 J/kg.K

m is mass of water, = 1.8 kg

Δθ is change in temperature, = 30.5 - 21 = 9.5 °C

Q = 1.8 x 4200 x  9.5

Q = 71820 J

Mass of nitrogen gas N₂, at this quantity of heat;

$m_{N_2} = \frac{Q}{C*\delta \theta} \\\\m_{N_2} = \frac{71820}{715.86*9.5}\\\\m_{N_2} = 10.56 \ kg$

The volume this air occupy at 21∘C

Apply ideal gas law;

$PV = nRT = \frac{m}{M} RT$

$PV = \frac{mRT}{M} \\\\V = \frac{mRT}{MP}\\\\V = \frac{10.56(kg)*8.314*10^3(L.Pa/mol.K)*294(K)}{29*10^{-3}(kg)1.01325*10^5 (Pa)}\\\\V = 8784.29 \ Liters$