- Home
- Standard 11
- Physics
11.Thermodynamics
medium
A certain amount of gas of volume $V$ at $27^{o}\,C$ temperature and pressure $2 \times 10^{7} \;Nm ^{-2}$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use $\gamma=1.5$ )
A
$3.536 \times 10^{5}\,Pa$
B
$3.536 \times 10^{6}\,Pa$
C
$1.25 \times 10^{6}\,Pa$
D
$1.25 \times 10^{5}\,Pa$
(JEE MAIN-2022)
Solution
$P _{1}=2 \times 10^{7} Pa$
$P _{1} V _{1}= P _{2} V _{2}$
Since $V_{2}=2 V_{1}$ Hence $P_{2}=P_{1} / 2$ (isothermal expansion)
$P _{2}=1 \times 10^{7} Pa$
$P _{2}\left( V _{2}\right)^{\gamma}= P _{3}\left(2 V _{2}\right)^{\gamma}$
$P _{3}=\frac{1 \times 10^{7}}{2^{1.5}}=3.536 \times 10^{6}$
Standard 11
Physics
