Respuesta :
The initial pressure in the cylinder of 10 bar of argon with [tex]C_p[/tex] = (5/2)·R
and the added gas at pressure of 50 bar gives the following values.
(a) 400.6 K
(b) 51.184 J
(c) The pressure of the cylinder when it is shipped is 37.194 bar
How can the temperature, heat transferred and pressure be calculated?
(a) Adiabatic compression, we have;
[tex]\dfrac{p_{1}}{p_{2}} = \mathbf{\left (\dfrac{T_{_{1}}}{T_{2}} \right )^{\dfrac{k}{k-1}}}[/tex]
Where;
[tex]k = \dfrac{C_p}{C_v} = \mathbf{\dfrac{C_p}{C_p - R}}[/tex]
Therefore;
[tex]K= \mathbf{\dfrac{\frac{5}{2} \cdot R}{\frac{5}{2} \cdot R - R} }= \dfrac{5}{3}[/tex]
Which gives;
[tex]\dfrac{10}{50} = \left (\dfrac{298}{T_{2}} \right )^{\dfrac{\frac{5}{3} }{\frac{5}{3} -1}} = \left (\dfrac{298}{T_{2}} \right )^{2.5}[/tex]
[tex]\dfrac{50}{10} = \mathbf{ \left (\dfrac{T_2}{298.15} \right )^{2.5}}[/tex]
[tex]ln(5)= 2.5\cdot ln\left (\dfrac{T_2}{298.15} \right )[/tex]
[tex]T_2 = 298 \times e^{\dfrac{ln(5)}{2.5} } \approx \mathbf{567.29\, K}[/tex]
The temperature
Using an ideal gas model, we have;
[tex]\dfrac{V_{1}}{V_{2}} = \mathbf{\dfrac{P_{2}\times T_{1}}{T_{2} \times P_1}}[/tex]
Which gives;
[tex]\dfrac{V_{1}}{V_{2}} = \dfrac{50\times 298}{567.29 \times 10} \approx 2.625[/tex]
V₁ = 2.625·V₂
Volume occupied by the gas in the cylinder after the pressure increases is therefore;
[tex]V_2 = \dfrac{V_1}{2.625} \approx 0.381 \cdot V_1[/tex]
Volume of gas added is therefore;
[tex]V =V_1 - \dfrac{V_1}{2.625} = 0.619\cdot V_1[/tex]
Considering a molar volume of gas, we have;
0.040 × 0.619 × 2.5×8.3145×(T₃ - 298) = 0.040 × 0.381 × 2.5×8.3145×(567.29 - T₃)
Solving gives;
T₃ ≈ 400.6 K
For a molar volume of gas cylinder, the temperature just after the valve is closed is T₃ ≈ 400.6 K
(b) If the cylinder sits in the storage for a long time, and heat is conducted out, we have;
The temperature of the cylinder will revert to room temperature of 298 K
The heat transferred is therefore;
0.040 × (2.5 × 8.3145 - 8.3145) ×(400.6 - 298) ≈ 51.184
The heat transferred from the cylinder is 51.184 J
(c) The pressure of the gas in the cylinder is therefore;
[tex]\dfrac{P_1}{T_1} = \mathbf{\dfrac{P_2}{T_2}}[/tex]
Which gives;
[tex]P_2= \mathbf{\dfrac{P_1}{T_1} \times T_2}[/tex]
Therefore;
[tex]P_2= \dfrac{50}{400.6} \times 298 \approx 37.194[/tex]
The pressure of the cylinder when it is shipped is 37.194 bar
Learn more about thermodynamics here:
https://brainly.com/question/5374634
