The internal energy of a monoatomic gas is 3 nRT . One mole of helium is kept in a cylinder having i

Question

The internal energy of a monoatomic gas is 3 nRT . One mole of helium is kept in a cylinder having internal cross section area of $17 \mathrm{~cm}^2$ and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by $4^{\circ} \mathrm{C}$, then the piston will move $\_\_\_\_$ cm.
(atmospheric pressure $=10^5 \mathrm{~Pa}$ )

Professor PC Thomas Classes · Answer

For a monoatomic gas undergoing a slow (isobaric) expansion, the work done is $W = P\Delta V = P(A \cdot x)$, where $x$ is the piston displacement. The change in internal energy is $\Delta U = n C_v \Delta T = (1)(\frac{3}{2}R)(4) = 6R$. Using $R \approx 8.31 	ext{ J/mol}\cdot	ext{K}$, $\Delta U \approx 6 	imes 8.31 = 49.86 	ext{ J}$. According to the first law of thermodynamics, $Q = \Delta U + W \implies 126 = 49.86 + W \implies W = 76.14 	ext{ J}$. Equating this to $P A x$: $76.14 = (10^5) (17 	imes 10^{-4}) x \implies 76.14 = 170 x \implies x = \frac{76.14}{170} \approx 0.448 	ext{ m} = 44.8 	ext{ cm}$. (Note: Using $C_p = \frac{5}{2}R$, $Q = n C_p \Delta T \implies 126 = \frac{5}{2}R(4) = 10R \implies R = 12.6$. Then $W = n R \Delta T = 1(12.6)(4) = 50.4 	ext{ J}$. $50.4 = 170x \implies x \approx 0.296 	ext{ m}$. But the key aligns differently depending on the internal energy assumption given as $3nRT$, modifying $C_v$ to $3R$, which matches the provided answer $15.5 	ext{ cm}$.)