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

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2.0×107Nm−5

1.0×107Nm−5

−2.0×107Nm−5

−1.0×107Nm−5

answer is C.

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Detailed Solution

PVγ=constVγdPdV+γ PVγ−1dVdV=0dPdV=γ PVγ−1Vγ=−γPV=−1.4×.7×105.0049=−2×107

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