Question

It is found on heating a gas,its volume increases by 50% and the pressure decreases to 60% of its original value. If the original temperature was -15°C, find the temperature to which it was healed.

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Answer 1

Question:-

➡ It is found on heating a gas,its volume increases by 50% and the pressure decreases to 60% of its original value. If the original temperature was -15°C, find the temperature to which it was healed.

Answer:-

➡ The required temperature is -40. 8°C

Solution:-

Let us assume that,

$\sf P_{1} = x$

$\sf V_{1} = y$

And given that,

$\sf T_{1} = -15\degree C = 273\degree - 15\degree = 258K$

Now,

$\sf P_{2} =60x/100$

$\sf =0.6x$

$\sf \implies P_{2} = 0.6x$

Now,

$\sf V_{2} = y + \frac{50y}{100}$

$\sf = \frac{150y}{100}$

$\sf = \frac{3y}{2}$

And,

$\sf T_{2}= ??$

Now,

we know that,

$\sf\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

So,

$\sf \implies \frac{xy}{258} = \frac{0.6x \times \frac{3y}{2} }{T _{2} }$

$\sf \implies \frac{1}{258} = \frac{0.9}{T _{2} }$

$\sf \implies T _{2} = 258 \times 0.9$

$\sf \implies T _{2} = 232.2K$

$\sf \implies T _{2} = (232.2 - 273) \degree C$

$\sf \implies T _{2} = - 40.8 \degree C$

Hence, the temperature at which it was healed is -40.8°C.