Chapter 10 - Temperature and Kinetic Theory - Learning Path Questions and Exercises - Exercises - Page 383: 34

$5.124cm^{3}$

As per ideal gas equation, $\frac{P1\times V1}{T1}=\frac{P2\times V2}{T2}$ $T1=380K$, $T2=393K$, $V1=2cm^{3}$, $P2=101300Pa$, $P1=101300+density \times gh=101300+1000\times 9.8\times 15=248300Pa$ So, solving for $V2$, we have $V2=5.124cm^{3}$