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. Show working please
To calculate the final temperature of the gas after it is compressed adiabatically, we can use the adiabatic equation:
\[ P_1V_1^{\gamma} = P_2V_2^{\gamma} \]
Where:
- \( P_1 \) is the initial pressure of the gas (760 mmHg)
- \( V_1 \) is the initial volume of the gas
- \( P_2 \) is the final pressure of the gas
- \( V_2 \) is the final volume of the gas
- \( \gamma \) is the adiabatic index, which is given by \( \gamma = \frac{C_p}{C_v} \)
In this case, the volume is halved, so \( V_2 = \frac{1}{2} V_1 \).
Let's calculate the adiabatic index first:
\[ \gamma = \frac{C_p}{C_v} = \frac{2100 \, \text{J/kg/K}}{1500 \, \text{J/kg/K}} = 1.4 \]
Now we can substitute the values into the adiabatic equation:
\[ 760 \, \text{mmHg} \times V_1^{1.4} = P_2 \times \left(\frac{1}{2} V_1\right)^{1.4} \]
Simplifying the equation further:
\[ 760 \, \text{mmHg} \times V_1^{1.4} = P_2 \times \left(\frac{1}{2}\right)^{1.4} \times V_1^{1.4} \]
Cancelling out \( V_1^{1.4} \) on both sides of the equation, we get:
\[ 760 \, \text{mmHg} = P_2 \times \left(\frac{1}{2}\right)^{1.4} \]
Now we solve for \( P_2 \):
\[ P_2 = \frac{760 \, \text{mmHg}}{\left(\frac{1}{2}\right)^{1.4}} \]
Using a calculator, we find \( P_2 \approx 1469.3 \) mmHg.
Now, we can use the ideal gas law to calculate the final temperature.
The ideal gas law states:
\[ PV = nRT \]
Where:
- \( P \) is the pressure of the gas (final pressure, \( P_2 \))
- \( V \) is the volume of the gas (final volume, \( \frac{1}{2} V_1 \))
- \( n \) is the number of moles of gas (assumed to be constant)
- \( R \) is the gas constant (8.314 J/mol/K)
- \( T \) is the temperature of the gas (final temperature, to be determined)
Rearranging the equation to solve for \( T \):
\[ T = \frac{PV}{nR} \]
Substituting the values into the equation:
\[ T = \frac{1469.3 \, \text{mmHg} \times \frac{1}{2} V_1}{n \times 8.314 \, \text{J/mol/K}} \]
Note: We need to convert the pressure from mmHg to pascals (Pa) and volume from any unit to cubic meters (m^3) to match the units of the gas constant.
Let's assume the number of moles is 1 for simplicity.
Converting mmHg to Pa, we have:
\[ 1 \, \text{mmHg} = 133.322 \, \text{Pa} \]
Converting volume to cubic meters:
\[ 1 \, \text{mL} = 10^{-6} \, \text{m^3} \]
\[ \frac{\frac{1}{2} V_1}{10^{-6} \, \text{m^3}} = \frac{1}{2} \times 10^6 \, V_1 \, \text{m^3} \]
Substituting the converted values into the equation, we get:
\[ T = \frac{1469.3 \times 133.322 \, \text{Pa} \times \frac{1}{2} \times 10^6 \, V_1 \, \text{m^3}}{1 \times 8.314 \, \text{J/mol/K}} \]
Simplifying the equation:
\[ T = \frac{768806.4 \, V_1 \, \text{m^3}}{8.314 \, \text{J/mol/K}} \]
Finally, the temperature can be calculated by substituting the given value of \( V_1 \), which is the initial volume of the gas.