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12.Kinetic Theory of Gases
normal
A gas at $27\,^oC$ has a volume $V$ and pressure $P$. On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be ...... $^oC$
A
$1800$
B
$162$
C
$1527$
D
$600$
Solution
From ideal gas equation $P V=\neq R T$
we get $\frac{T_{2}}{T_{1}}=\left(\frac{P_{2}}{P_{1}}\right)\left(\frac{V_{2}}{V_{1}}\right)=\left(\frac{2 P_{1}}{P_{1}}\right)\left(\frac{3 V_{1}}{V_{1}}\right)=6$
$\therefore T_{2}=6 T_{1}=6 \times 300=1800 K=1527^{\circ} \mathrm{C}$
Standard 11
Physics
