A certain volume of gas at 298k is heated such that its volume and pressure are now four times the original values. What is the new temperature?

Assessment: JAMB UTME - Chemistry - 1985 Subject: Chemistry

Question 1 Report

18.6k

100.0k

298.0k

1192.0k

4768.0k

Answer Details

\frac{P_{1} V_{1}}{T_{1}} = \frac{4 P_{1} \times 4 V_{1}}{T_{2}}

i.e $\frac{P_{1} V_{1}}{298} = \frac{4 P_{1} \times 4 V_{1}}{T_{2}}$

T2 = $\frac{298 \times 4 P_{1} \times 4 V_{1}}{P_{1} V_{1}}$

= 298 + 16 = 4768.0k

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