Question

The internal energy of air given, at ordinary temperature by,
$u=u_0+0.718t$
where u is in kJ/kg, $u_0$ is any arbitrary value of u at $0^{\circ }C$ in kJ/kg and t is temperature in ${}^{\circ }C$.
Also for air, $pv=0.287(t+273)$

where $\rho$ is in kPa and 0 is in $m^3/kg$.
An evacuated bottle is fitted with a value through which air from the atmosphere, at 760 mm Hg and ${25}^{\circ }C$ is allowed to flow slowly to fill the bottle. If no heat is transferred to or from the air in the bottle. What will be the temperature when the pressure in the bottle reaches 760 mm Hg?

• A. 298 K
• B. 387 K
• C. 417.2 K
• D. 432 K

Answer 1

The correct option is C 417.2 K

As per given information

As we know from mass conserved on

$({\dot{m}}_2-{\dot{m}}_1)=({\dot{m}}_i-{\dot{m}}_e)$

Since, ${\dot{m}}_e=0$

${\dot{m}}_i=({\dot{m}}_2-{\dot{m}}_1)$

From unsteady state analysis,

$\frac{du}{dt}={\dot{m}}_i(h_i+\frac{V_i^2}{2}+g{Z}_i)+Q_{cu}-{\dot{m}}_e(h_e+\frac{V_e^2}{2}+g{Z}_e)-W_{cu}$

$\frac{du}{dt}={\dot{m}}_i{h}_i$

$du={\dot{m}}_i{h}_{i}dt$

$du=\left(\frac{dm}{dt}{h}_i\right)dt$

$m_2{u}_2-m_i{u}_i=(m_2-m_2)h_1$

$T_2=\gamma T_1\Rightarrow T_2=(25+273)\times 1.4=417.2K$