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← JEE Main 2014 · 11 Apr · Shift 1
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JEE Main 2014 · 11 Apr · Shift 1, Physics question 13

An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross section 8.0×10−38.0\times10^{-3} m2^2. Initially the gas is at 300K and occupies a volume of 2.4×10−32.4\times10^{-3} m3^3 and the spring is in its relaxed state as shown in figure. The gas is heated by a small heater until the piston moves out slowly by 0.1 m. The force constant of the spring is 8000 N/m and the atmospheric pressure is 1.0×1051.0\times10^5 N/m2^2. The cylinder and the piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and the cylinder. The final temperature of the gas will be :

(Neglect the heat loss through the lead wires of the heater. The heat capacity of the heater coil is also negligible)

A horizontal cylinder with a heater coil inside on the left; a piston on the right is attached to a horizontal spring whose other end is fixed to a wall.
Choose one option
Answer and worked solution

Official answer: B

The spring compresses by 0.1 m, adding 8000×0.18×10−3=105\dfrac{8000\times0.1}{8\times10^{-3}}=10^5 Pa. So P2=2×105P_2=2\times10^5 Pa, and
V2=2.4×10−3+0.1×8×10−3=3.2×10−3 m3.V_2=2.4\times10^{-3}+0.1\times8\times10^{-3}=3.2\times10^{-3}\text{ m}^3.
T2=300×2×105×3.2×10−3105×2.4×10−3=800 K.T_2=300\times\frac{2\times10^5\times3.2\times10^{-3}}{10^5\times2.4\times10^{-3}}=800\text{ K}.

Topics:Ideal gas equationThermodynamic processes
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