91Ó°ÊÓ

Given data;

Mass of the nitrogen$M=28×{10}^{-3}kg/mol$

Specific heat of water$c_{water}=4190J/molK$

Temperature difference $∆T=10.0K$

Presser is $P=1atm=1.013×{15}^{5}Pa$

Gas constant$R=8.314J/molK$

Assumption;

Air is 100% N₂

Equation;

Specific heat capacity of nitrogen

$c=\frac{5R}{3M}$ .......... (1)

The amount of heat that is warm up

$Q={\mathrm{mc}}_{\mathrm{water}}∆\mathrm{T}$ .......... (2)

The mass of the warmed nitrogen

${\mathrm{m}}_{{\mathrm{n}}_2}=\frac{\mathrm{Q}}{{\mathrm{c}}_{{\mathrm{n}}_2}∆\mathrm{T}}$ .......... (3)

The ideal gas law

$V=\frac{nRT}{P}=\frac{mRT}{MP}$ .......... (4)

(a)

Put values in equation (1)

$\begin{array}{r}\mathrm{c}=\frac{5}{2}\left(\frac{8.314\mathrm{J}/\mathrm{molK}}{24×{10}^{−3}\mathrm{kg}/\mathrm{mol}}\right)\\ =742.3\mathrm{J}/\mathrm{kg}â‹\dots \mathrm{K}\end{array}$

Hence, the specific heat of nitrogen is$742.3\mathrm{J}/\mathrm{kg}â‹\dots \mathrm{K}$

(b)

Put values in equation (2)

$$\begin{gathered} Q=\left(1.00kg\right)\left(4190J/molK\right)\left(10.0K\right) \\ =4.19×{10}^{4}J \end{gathered}$$

Put this value in equation (3)

$\begin{array}{r}{\mathrm{m}}_{{\mathrm{n}}_2}=\frac{4.19×{10}^4\mathrm{J}}{(742.3\mathrm{J}/\mathrm{kg}K\mathrm{K})\left(10.0\mathrm{K}\right)}\\ =5.64\mathrm{kg}\end{array}$

Hence, the mass of warmed nitrogen isrole="math" localid="1668187245250" $m_{n_2}=5.6\mathrm{kg}$

Put the all the values in equation (4)

$\begin{array}{r}V=\frac{\left(5.65\mathrm{kg}\right)(8.314\mathrm{J}/\mathrm{molK})\left(293\mathrm{K}\right)}{\left(28.014×{10}^{−3}\mathrm{kg}/\mathrm{mol}\right)\left(1.013×{10}^5\mathrm{Pa}\right)}\\ =4.85{\mathrm{m}}^3\end{array}$

Hence, the volume of warmed nitrogen is $\mathrm{V}=4.85m^3$