During a compression process involving an ideal gas at pressure P\(_1\), when the volume, V\(_1\)...
During a compression process involving an ideal gas at pressure P\(_1\), when the volume, V\(_1\) of the gas was halved, the temperature in Kelvin increases by half its initial value. The final pressure P\(_2\) is given by
- A)3P\(_1\)
- B)12P\(_1\)
- C)6P\(_1\)
- D)1.5P\(_1\)
Correct Answer: A)3P\(_1\)
Explanation
In order to explain this question, we should first recall the ideal gas law, which relates the pressure, volume, and temperature of an ideal gas. The ideal gas law is given by the equation:
\(PV = nRT\)
Where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin.
In this question, we are given that the volume of the gas is halved, which means that the final volume V\(_2\) is half of the initial volume V\(_1\). Mathematically, this can be represented as:
\(V_2 = \frac{1}{2}V_1\)
Additionally, we are given that the temperature increases by half its initial value. Therefore, the final temperature T\(_2\) is 1.5 times the initial temperature T\(_1\). Mathematically, this is:
\(T_2 = 1.5T_1\)
Now, we can use the ideal gas law to find the final pressure P\(_2\). Since the number of moles and the ideal gas constant remain constant, we can write the initial and final states of the gas as:
\(P_1V_1 = nRT_1\) and \(P_2V_2 = nRT_2\)
Divide the two equations to eliminate nR:
\(\frac{P_2V_2}{P_1V_1} = \frac{T_2}{T_1}\)
Substitute the expressions for V\(_2\) and T\(_2\):
\(\frac{P_2(\frac{1}{2}V_1)}{P_1V_1} = \frac{1.5T_1}{T_1}\)
We can see that V\(_1\) and T\(_1\) cancel out:
\(\frac{P_2}{2P_1} = 1.5\)
Finally, isolate P\(_2\) to find the final pressure:
\(P_2 = 3P_1\)
Therefore, the correct answer isOption A: 3P\(_1\).