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Dimensional Analysis

An ideal mono...

Question

An ideal mono-atomic gas is confined in a cylinder by a spring loaded piston of cross-section $8 \times 10^{- 3} m^{2}$ . Initially the gas is at 300k and occupies a volume of $2.4 \times 10^{- 3} m^{3}$ and the spring is in its relaxed (upstretched, uncompressed) state. The gas is heated by a small electric heater until the piston moves out slowly by 0.1m. The force constant of spring is 8000 N/m and the atmospheric pressure is $1 \times 10^{5} N / m^{2}$ . The heat supplied in joules by the heater is [Consider the cylinder and the piston as thermally insulated, the piston and spring to be massless, friction to be absent, negligible heat loss through lead wires of the heater and negligible heat capacity of the heater coil.]

600 J

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720 J

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800 J

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120 J

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Solution

The correct option is B 720 J
$\frac{P V}{T} = c o n s t a n t \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} P_{1} = P_{a t m} = 10^{5} N / m^{2} V_{1} = 2.4 \times 10^{- 3} m^{3} T_{1} = 300 K$
Balancing forces on the piston, we get
$P_{2} = P_{a t m} + \frac{k x}{A}$
$\Rightarrow P_{2} = 10^{5} + 10^{5} = 2 \times 10^{5} N / m^{2}$ and $V_{2} = V_{1} + x A = 3.2 \times 10^{- 3} m^{3}$
$\Rightarrow T_{2} = (\frac{P_{2} V_{2}}{P_{1} V_{1}}) T_{1} = 800 K$
Now, $Δ Q = Δ U + W$
Where \Delta U=nC_v \Delta T\)
$= \frac{P_{1} V_{1}}{R T_{1}} \times (\frac{3}{2} R) \times (T_{2} - T_{1}) = \frac{10^{5} \times 2.4 \times 10^{- 3} \times 3}{300 \times 2} \times 500 = 600 J$
W = work done against atmosphere + work done against spring
$W = P_{a t m} Δ V + \frac{1}{2} k x^{2} = 10^{5} (0.1 \times 8 \times 10^{- 3}) + 0.5 \times 8000 \times 0.1$
$W = 120 J$
Hence, $Δ Q = Δ U + W$
$= 600 + 120 = 720 J$ .

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