Given: \( m=2\,\text{kg} \), \( T_{1}=30^{\circ}\text{C}=303\,\text{K} \), \( p_{1}=100\,\text{kPa} \), isentropic process, \( T_{2}=850^{\circ}\text{C}=1123\,\text{K} \).

To calculate: \( p_{2}=? \), \( \rho_{1}=? \), \( \rho_{2}=? \), \( v_{1}=? \), \( v_{2}=? \).

Using isentropic relation, with \( \gamma=1.4 \) for air,

$$ \frac{p_{2}}{p_{1}}=\left(\frac{T_{2}}{T_{1}}\right)^{\gamma/\left(\gamma-1\right)} $$ $$ p_{2}=p_{1}\left(\frac{T_{2}}{T_{1}}\right)^{\gamma/\left(\gamma-1\right)} $$ $$ p_{2}=100\times10^{3}\times\left(\frac{1123}{303}\right)^{1.4/\left(1.4-1\right)} $$ $$ \boxed{p_{2}=9801.216\,\text{kPa}}\ . $$

The density \( \rho_{1} \) can be calculated using the ideal gas equation as,

$$ \rho_{1}=\frac{p_{1}}{R\,T_{1}}=\frac{100\times10^{3}}{287\times303} $$ $$ \boxed{\rho_{1}=1.15\,\text{kg/m}^{3}}\ . $$

The density \( \rho_{2} \) can be calculated using the ideal gas equation as,

$$ \rho_{2}=\frac{p_{2}}{R\,T_{2}}=\frac{9801.216\times10^{3}}{287\times1123} $$ $$ \boxed{\rho_{2}=30.41\,\text{kg/m}^{3}}\ . $$

Since the mass is constant \( m=2\,\text{kg} \), the volumes can be calculated as,

$$ \boxed{v_{1}=\frac{m}{\rho_{1}}=\frac{2}{1.15}=1.73913\,\text{m}^{3}} $$ $$ \boxed{v_{2}=\frac{m}{\rho_{2}}=\frac{2}{30.41}=0.0657678\,\text{m}^{3}} $$