A cylinder contains acetylene gas at $27^{0} C$ and 4.05 M Pa. The pressure in the cylinder after half the mass of gas is used up and temperature has fallen to $12^{0} C$ will be

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4.05 MPa

2.025 MPa

3.84 MPa

1.92 MPa

answer is D.

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Detailed Solution

$PV = \frac{w}{M} RT \frac{P}{wT} = constant$

$\begin{array}{l} \frac{P_{1}}{w_{1} T_{1}} = \frac{P_{2}}{w_{2} T_{2}} ⇒ \frac{4 .05}{w \times 300} = \frac{P_{2}}{\frac{w}{2} \times 285} \\ P_{2} = 1 .92 MPa \end{array}$

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