Question

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the pressure of a monatomic ideal gas is doubled, while the volume is cut in half. by what factor is the internal energy of the gas multiplied?

○ 4
○ 1
○ \\(\frac{1}{2}\\)
○ \\(\frac{1}{4}\\)
○ 2

Explanation:

Step1: Recall Ideal Gas Law and Internal Energy Formula

The ideal gas law is \( PV = nRT \), and the internal energy \( U \) of a monatomic ideal gas is \( U=\frac{3}{2}nRT \). From the ideal gas law, \( nRT = PV \), so we can substitute into the internal energy formula: \( U=\frac{3}{2}PV \).

Step2: Analyze Initial and Final States

Let the initial pressure be \( P_1 \), initial volume be \( V_1 \), so initial internal energy \( U_1=\frac{3}{2}P_1V_1 \).

Final pressure \( P_2 = 2P_1 \) (doubled), final volume \( V_2=\frac{1}{2}V_1 \) (cut in half). Then final internal energy \( U_2=\frac{3}{2}P_2V_2 \).

Substitute \( P_2 \) and \( V_2 \) into the formula: \( U_2=\frac{3}{2}(2P_1)(\frac{1}{2}V_1) \).

Step3: Simplify the Expression for \( U_2 \)

Simplify \( \frac{3}{2}(2P_1)(\frac{1}{2}V_1) \): the 2 and \( \frac{1}{2} \) cancel out, so \( U_2=\frac{3}{2}P_1V_1 \). But \( U_1=\frac{3}{2}P_1V_1 \), so \( U_2 = U_1 \). This means the internal energy is multiplied by a factor of 1.

Answer:

1 (corresponding to the option "1")