It is found that when a certain gas is heated, its volume increases by $50$ % and pressure decreases to $75$ % of its original value. If the original temperature of the gas was $- 15^{o} C$ , find the temperature(in Celsius) to which of it was heated:

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Solution

Initial volume= $V_{1}$ , pressure= $P_{1}$ and temperature= $T_{1}$ .
Final Volume= $V_{2} = \frac{150}{100} = 1.5 V_{1}$ as volume increases by 50%.
Final Pressure= $P_{2} = 0.75 P_{1}$ as pressure decreases to 75% of its original pressure.
Original temperature= $T_{1}$ =-15 $^{\circ}$ C=258 K
Now, $\frac{P V}{T} = n R$ = constant for a particular gas having 1 mole.
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$⟹ \frac{P_{1} V_{1}}{258} = \frac{{0.75 P}_{1} \times 1.5 V_{1}}{T_{2}} ⟹ T_{2} =$ 290.25 K=17.25 $^{\circ}$ C

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It is found that when a certain gas is heated, its volume increases by 50% and pressure decreases to 75% of its original value. If the original temperature of the gas was −15oC, find the temperature(in Celsius) to which of it was heated:

It is found that when a certain gas is heated, its volume increases by $50$ % and pressure decreases to $75$ % of its original value. If the original temperature of the gas was $- 15^{o} C$ , find the temperature(in Celsius) to which of it was heated: