WAEC - Chemistry (1990 - No. 25)
To what temperature must a gas be raised from 273K in order to double both its volume and pressure?``
298k
300k
546k
819k
1092k
Explanation
Given:
T\(_1\) = 273K V\(_1\) = X P\(_1\) = Y → Let x and y represent the initial volume and initial pressure respectively.
T\(_2\) = ? V\(_2\) = 2X P\(_2\) = 2Y → Since the volume and pressure were doubled.
Using the formula, \(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2 V_2}{T_2}\)
T\(_2\) = \(\frac{P_2 V_2 T_1}{P_1 V_1}\)
= \(\frac{2Y\times 2X \times 273}{Y\times X}\)
= 2 X 2 X 273
T\(_2\)= 1092K
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