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JEE Advanced levelSingle correct optionDifficulty 3 of 3ig095

The density of a gas is $d_{1}$ at temperature $T$ and pressure $P$ . If the pressure is doubled and the kelvin temperature is reduced by one third, the new density is

A $3 d_{1}$
B $\frac{4 d _{1}}{3}$
C $\frac{3 d _{1}}{2}$
D $6 d_{1}$

Answer: A

d \propto P / T

. New

P = 2 P

, new

T = T - T /3 = 2 T /3

.

d_{2} = d_{1} \times \frac{2}{2/3} = 3 d_{1}

.

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