Air is held inside a non-insulated cylinder using a piston mass ( $M = 25 k g$ and area $A = 100 c m^{2}$ ) and stoppers (of negligible area), as shown in the figure. The initial pressure $P_{1}$ and temperature $T_{1}$ of air inside the cylinder are $200 k P a$ and $400^{0} C$ , respectively. The ambient pressure $P_{\infty}$ and temperature $T_{\infty}$ are $100 k P a$ and $27^{0} C$ respectively. The temperature of the air inside the cylinder $(^{0} C)$ at which the piston will begin to move is (correct to two decimal places).

147.625

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Solution

The correct option is A 147.625

$P_{\infty} = 100 k P a$

$P_{p i s t o n} = \frac{M g}{A} = \frac{25 \times 10 \times 10^{- 3}}{0.01}$
$P_{p i s t i o n} = 25 k P a$
Total external pressure
$= P_{\infty} + P_{p i s t i o n} = 125 k P a$ .
Initial pressure of air,
$P_{i} = 200 k P a$ .
Piston will be about to come down if system pressure is equal to total pressure i.e 125 kPa. $\Rightarrow$ Final pressure of air at which it is about to move is $P_{f} = 125 k P a$
Applying mass conservation at two states i and f

$m_{i} = m_{f}$
$\frac{P_{i} V_{i}}{.} R T_{i} = \frac{P_{f} V_{f}}{R T_{f}} [∵ V_{i} = V_{f}]$
$\frac{P_{i}}{T_{i}} = \frac{P_{f}}{T_{f}}$
$\frac{200}{673} = \frac{125}{T_{f}}$

$T_{f} = 420.625 k$
$T_{f} = {147.625}^{0} C$

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