An ideal monatomic gas is confined in a cylinder by a spring loaded piston of cross section 8 x10^-3m² . Initially the gas is at 300K and occupies a volume of 2.4 x 10^-3 m³ and the spring is in its relaxed position. The gas is heated by a small electric heater until the piston moves out slowly by 0.1m. Calculate the heat supplied (in J) by the heater. The force constant of the spring is 8000 N/m and atmospheric pressure 1x10⁵N / m²

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answer is 720.

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Detailed Solution

As here initially
$\begin{array}{l} P_{1} = P_{0} = 1 \times 10^{5} N / m^{2} \\ V_{1} = V_{0} = 2 .4 \times 10^{- 3} m^{3} \\ T_{1} = T_{0} = 300 K and finally \\ P_{F} = P_{0} + \frac{Kx}{A} = 1 \times 10^{5} + \frac{8000 \times 0 .1}{8 \times 10^{- 3}} = 2 \times 10^{5} N / m^{2}, \\ V_{F} = V_{0} + Ax = 2 .4 \times 10 + 0 .1 \times 8 \times 10^{- 3} = 3 .2 \times 10^{- 3} m^{3} \end{array}$
So from gas equation $PV = μRT, i.e. \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{F} V_{F}}{T_{F}}$
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we have $T_{F} = \frac{P_{F}}{P_{1}} \times \frac{V_{F}}{V_{1}} \times T_{1} = \frac{2 \times 10^{5}}{1 \times 10^{5}} \times \frac{3 .2 \times 10^{- 3}}{2 .4 \times 10^{- 3}} \times 300 = 800 K$
As $ΔW = \int PdV, Here P = P_{0} + (Kx / A)$
So $ΔW = \int_{0}^{0 .1} (P_{0} + \frac{kx}{A}) Adx = \int_{0}^{0 .1} (P_{0} A + kx) dx i.e. = P_{0} Ax + \frac{1}{2} {kx}^{2}$
$\begin{array}{l} = [10^{5} \times 8 \times 10^{- 3} \times 0 .1 + \frac{1}{2} \times 8000 \times (0 .1)^{2}] \\ i.e. ΔW = 80 + 40 = 120 J \\ ΔU = {nc}_{v} ΔT, ΔU = 600 J, ΔQ = ΔU + ΔW = 600 + 120 = 720 J \end{array}$