Q.
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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a
2.0×107Nm−5
b
1.0×107Nm−5
c
−2.0×107Nm−5
d
−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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