Question

At constant temperature, the volume of a gas is to be decreased by 4%. The pressure must be increased by

• A. 0.04
• B. 0.0416
• C. 0.08
• D. 0.0386

Answer 1

Answer: (B)

0.0416

Answer 2

At constant temperature, $p_1 V_1 = p_2 V_2$ $\Rightarrow \frac{ p_1 }{ p_2 } = \frac{ V_2 }{ V_1 }$ Hence fractional change in volume $\Rightarrow \frac{ V_1 - V_2 }{ V_1 } = \frac{ 4 }{ 100} = \frac{ 1}{ 25 }$ $\Rightarrow 1 - \frac{ V_2 }{ V_1 } = \frac{ 1}{25}$ $\Rightarrow \frac{ V_2 }{ V_1 } = \frac{ 24}{25}$ $\Rightarrow \frac{ p_1 }{ p_2 } = \frac{ V_2 }{ V_1 } = \frac{ 24}{25}$ $\Rightarrow \frac{ p_2 - p_! }{ p_1 } = \frac{ 24}{25} - 1 = \frac{ 1}{ 24 }$ Percentage increase in pressure = $\frac{ 100}{ 24} =$ 4.16%.