A steel tank contain air at a pressure of 15 bar at $20^{0} C .$ The tank is provided with a safety valve which can withstand a pressure of 30 bar. Calculate the temperature to which the tank can be safety heated.

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$313^{0} C$

$586^{0} C$

$- 313^{0} C$

313K

answer is B.

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Detailed Solution

Let $\begin{array}{l} P_{1} = 15 b a r, P_{2} = 30 b a r \\ T_{1} = 20^{0} C = 293 K, T_{2} = ? \end{array}$
Now, $\frac{P_{1}}{T_{1}} = \frac{P_{1}}{T_{2}} o r T_{2} = \frac{P_{2} T_{1}}{P_{1}} = \frac{30 \times 293}{15}$
$∴ T_{2} = 586 K o r 313^{0} C$
Thus, the safety valve will blow up when the temperature exceeds $313^{0} C$ .