an ideal gas at 17•c has a
pressure of 760 mmHg and is compressed
adiabatically until its volume is
halved.Calculate final temperature of the gas
assuming Cp= 2100 J/kg/K and Cv=1500J/kg/K
Solution
I need the solution for this question pls
To find the final temperature of the gas, we can use the adiabatic compression formula:
P1 * V1^γ = P2 * V2^γ
where P1 and V1 are the initial pressure and volume of the gas, P2 and V2 are the final pressure and volume of the gas, and γ is the ratio of specific heat capacities (Cp/Cv).
We are given:
Initial temperature (T1) = 17°C = 17 + 273 = 290 K
Initial pressure (P1) = 760 mmHg
Final volume (V2) = V1/2 = 0.5 * V1
First, let's find the initial volume (V1):
We know that the ideal gas equation is given by:
PV = nRT
Assuming the number of moles (n) is constant, we can rearrange the equation to solve for volume (V):
V = nRT / P
Since we're dealing with an ideal gas, the number of moles (n) is not given. However, we can assume it to be 1 for simplicity. So, we can calculate the initial volume (V1) as follows:
V1 = (1 * 8.314 * 290) / 760 = 3.169 m^3
Next, let's calculate the final pressure (P2):
Using the adiabatic compression formula, we have:
P1 * V1^γ = P2 * V2^γ
Substituting the given values, we get:
760 * (3.169)^γ = P2 * (0.5 * 3.169)^γ
Now let's calculate the value of γ:
γ = Cp / Cv = 2100 / 1500 = 1.4
Simplifying the equation further:
760 * (3.169)^1.4 = P2 * (0.5 * 3.169)^1.4
Solving this equation, we find:
P2 = 760 * (0.5 * 3.169)^(1.4/1.4) = 268.987 mmHg
Finally, let's calculate the final temperature (T2):
We can use the ideal gas equation again to find the final temperature:
P2 * V2 = nRT2
Simplifying this equation:
P2 * (0.5 * V1) = nRT2
Solving for T2:
T2 = (P2 * V2) / (nR)
Substituting the values and solving:
T2 = (268.987 mmHg * (0.5 * 3.169 m^3)) / (1 * 8.314 J/(mol K)) = 61.378 K
Therefore, the final temperature of the gas after adiabatic compression is approximately 61.378°C.
To calculate the final temperature of the gas after it is compressed adiabatically, we can use the adiabatic compression equation:
\(T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\frac{\gamma -1}{\gamma}}\)
Where:
\(T_1\) is the initial temperature of the gas
\(V_1\) is the initial volume of the gas
\(T_2\) is the final temperature of the gas
\(V_2\) is the final volume of the gas
\(\gamma\) is the heat capacity ratio (\(\gamma = \frac{C_p}{C_v}\))
Given:
\(T_1 = 17°C = 17 + 273 = 290K\)
\(P_1 = 760 mmHg\)
\(V_1 = V\)
\(V_2 = \frac{V}{2}\)
\(C_p = 2100 J/kg/K\)
\(C_v = 1500 J/kg/K\)
First, we need to convert the pressure to SI units (Pascal):
\(P_1 = 760 mmHg = 760 \times 133.322 Pa = 101325 Pa\)
Next, we need to find the value of \(\gamma\):
\(\gamma = \frac{C_p}{C_v} = \frac{2100}{1500} = 1.4\)
Now, we can substitute the given values into the adiabatic compression equation:
\(T_2 = 290 \times \left(\frac{V}{\frac{V}{2}}\right)^{\frac{1.4 - 1}{1.4}}\)
Simplifying this equation gives:
\(T_2 = 290 \times \left(\frac{2}{1}\right)^{\frac{0.4}{1.4}}\)
Now, we can calculate the final temperature of the gas:
\(T_2 = 290 \times 2^{0.286}\)
Using a calculator, we find:
\(T_2 \approx 290 \times 1.218\)
\(T_2 \approx 353.22K\)
Therefore, the final temperature of the gas after adiabatic compression is approximately 353.22K.