A certain amount of gas of volume V at 27° C temperature and pressure 2 × 107 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 γ=1.5 )

Question

A certain amount of gas of volume V at 27° C temperature and pressure 2 × 107 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 × 105 Pa (B) 3.536 × 106 Pa (C) 1.25 × 106 Pa (D) 1.25 × 105 Pa

Tardigrade Admin · Accepted Answer

P 1=2 × 107 Pa P 1 V 1= P 2 V 2 Since V2=2 V1 Hence P2=P1 / 2 (isothermal expansion) P2=1 × 107 Pa P2(V2)&gamma;=P3(2 V2)&gamma; P3=(1 × 107/21.5)=3.536 × 106

Q. A certain amount of gas of volume $V$ at $2 7^{\circ} C$ temperature and pressure $2 \times 1 0^{7} N m^{- 2}$ 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 $γ = 1.5$ )

$3.536 \times 1 0^{5} P a$

$3.536 \times 1 0^{6} P a$

$1.25 \times 1 0^{6} P a$

$1.25 \times 1 0^{5} P a$

$P_{1} = 2 \times 1 0^{7} P a$
$P_{1} V_{1} = P_{2} V_{2}$
Since $V_{2} = 2 V_{1}$ Hence $P_{2} = P_{1} /2$ (isothermal expansion)
$P_{2} = 1 \times 1 0^{7} P a$
$P_{2} (V_{2})^{γ} = P_{3} (2 V_{2})^{γ}$
$P_{3} = \frac{1 \times 1 0 ^{7}}{2 ^{1.5}} = 3.536 \times 1 0^{6}$

Q. A certain amount of gas of volume V at 27∘C temperature and pressure 2×107Nm−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 γ=1.5 )

3.536×105Pa

3.536×106Pa

1.25×106Pa

1.25×105Pa