Question

current attempt in progress
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?
○ 2
○ \\(\frac{1}{4}\\)
○ 1
○ \\(\frac{1}{2}\\)
○ 4

Explanation:

Step1: Recall Ideal Gas Law and Internal Energy Formula

The ideal gas law is \( PV = nRT \), and the internal energy of a monatomic ideal gas is \( U=\frac{3}{2}nRT \) (for monatomic, the degree of freedom \( f = 3 \), so \( U=\frac{f}{2}nRT=\frac{3}{2}nRT \)). We can also express \( U \) in terms of \( P \) and \( V \) using \( PV = nRT \), so \( U=\frac{3}{2}PV \).

Step2: Analyze Initial and Final States

Let 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).

Final internal energy \( U_2=\frac{3}{2}P_2V_2 \). Substitute \( P_2 = 2P_1 \) and \( V_2=\frac{1}{2}V_1 \):

\( U_2=\frac{3}{2}(2P_1)(\frac{1}{2}V_1) \)

Step3: Simplify the Expression for \( U_2 \)

Simplify \( (2P_1)(\frac{1}{2}V_1)=P_1V_1 \). So \( U_2=\frac{3}{2}P_1V_1 \). But \( U_1=\frac{3}{2}P_1V_1 \), so \( U_2 = U_1 \). The factor is \( \frac{U_2}{U_1}=1 \).

Answer:

1 (corresponding to the option "1")