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Question
It is found that on heating a gas its volume increases by 50% and its pressure decreases to 60% of its original value. If the original temperature was −15°C, find the temperature to which it was heated.
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Solution
Let the original volume (V) = 1 and
the original pressure (P) = 1 and
the temperature given (T) = −15°C.
= −15 + 273
= 258 K
V1 or new volume after heating = original volume + 50% of original volume
= `1 + 1 xx 50/100`
= `1 + 1/2`
= `3/2`
P1 or decreased pressure = 60%
`1 xx 60/100 = 0.6`
T1 = to be calculated
`(PV)/T = (P_1 V_1)/T_1`
`(1 xx 1)/258 = 3/2 xx 0.6/T_1`
T1 = 232.2
T1 = 232.2 − 273
= −40.8°C
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