A reversible adiabatic path on a P-V diagram for an ideal gas passes through state A where P =0.7 × 105 N / m -2 and V =0.0049 m 3. The ratio of specific heat of the gas is 1.4. The slope of path at A is

Question

A reversible adiabatic path on a P-V diagram for an ideal gas passes through state A where P =0.7 × 105 N / m -2 and V =0.0049 m 3. The ratio of specific heat of the gas is 1.4. The slope of path at A is (A) 2.0 × 107 Nm -5 (B) 1.0 × 107 Nm -5 (C) -2.0 × 107 Nm -5 (D) -1.0 × 107 Nm -5

Tardigrade Admin · Accepted Answer

PV &gamma;= constant V &gamma; ( d P / dV )=&gamma; PV &gamma;-1 ( d V / dV )=0 ( dP / dV )=(-&gamma; PV &gamma;-1/ V &gamma;)=(-&gamma; P / V ) =-1.4 × (0.7 × 105/0.0049) =-2 × 107

Q. A reversible adiabatic path on a $P - V$ diagram for an ideal gas passes through state $A$ where $P = 0.7 \times 1 0^{5} N / m^{- 2}$ and $V = 0.0049 m^{3}$ . The ratio of specific heat of the gas is $1.4$ . The slope of path at $A$ is

$2.0 \times 1 0^{7} N m^{- 5}$

$1.0 \times 1 0^{7} N m^{- 5}$

$- 2.0 \times 1 0^{7} N m^{- 5}$

$- 1.0 \times 1 0^{7} N m^{- 5}$

$P V^{γ} =$ constant
$V^{γ} \frac{d P}{d V} = γ P V^{γ - 1} \frac{d V}{d V} = 0$
$\frac{d P}{d V} = \frac{- γ P V ^{γ - 1}}{V ^{γ}} = \frac{- γ P}{V}$
$= - 1.4 \times \frac{0.7 \times 1 0 ^{5}}{0.0049}$
$= - 2 \times 1 0^{7}$

Q. A reversible adiabatic path on a P−V diagram for an ideal gas passes through state A where P=0.7×105N/m−2 and V=0.0049m3. The ratio of specific heat of the gas is 1.4. The slope of path at A is

2.0×107Nm−5

1.0×107Nm−5

−2.0×107Nm−5

−1.0×107Nm−5