Solution 25: Maximum Temperature in Piston Compression
Reasoning:
Maximum compression occurs when relative velocity is zero. Both pistons move at center of mass velocity $v_{cm} = u/2$. Loss in kinetic energy is converted to internal energy. $$ \Delta KE_{\text{loss}} = \frac{1}{2}mu^2 – \frac{1}{2}(2m)\left(\frac{u}{2}\right)^2 = \frac{1}{4}mu^2 $$
Maximum compression occurs when relative velocity is zero. Both pistons move at center of mass velocity $v_{cm} = u/2$. Loss in kinetic energy is converted to internal energy. $$ \Delta KE_{\text{loss}} = \frac{1}{2}mu^2 – \frac{1}{2}(2m)\left(\frac{u}{2}\right)^2 = \frac{1}{4}mu^2 $$
Calculation:
Helium is monoatomic ($C_v = \frac{3}{2}R$). $$ \Delta U = n C_v \Delta T = \Delta KE_{\text{loss}} $$ $$ n \left( \frac{3}{2}R \right) \Delta T = \frac{1}{4}mu^2 $$ $$ \Delta T = \frac{m u^2}{6 n R} $$ Substituting values ($m = 0.415$ kg, $u = 12$ m/s, $n = 0.1$, $R \approx 8.31$): $$ \Delta T = \frac{0.415 \times 144}{0.6 \times 8.31} \approx \frac{59.76}{4.986} \approx 12 \text{ K} $$ Answer: The maximum temperature change is 12 K.
Helium is monoatomic ($C_v = \frac{3}{2}R$). $$ \Delta U = n C_v \Delta T = \Delta KE_{\text{loss}} $$ $$ n \left( \frac{3}{2}R \right) \Delta T = \frac{1}{4}mu^2 $$ $$ \Delta T = \frac{m u^2}{6 n R} $$ Substituting values ($m = 0.415$ kg, $u = 12$ m/s, $n = 0.1$, $R \approx 8.31$): $$ \Delta T = \frac{0.415 \times 144}{0.6 \times 8.31} \approx \frac{59.76}{4.986} \approx 12 \text{ K} $$ Answer: The maximum temperature change is 12 K.