Chapter 3: Q9E (page 610)

A large cylindrical tank contains $0.750 m^{3}$ of nitrogen gas at role="math" localid="1668235464848" $27 ° C$ and $7.50 * 10^{3} P a$ (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to $0.410 m^{3}$ and the temperature is increased to $157 ° C$ ?

Short Answer

Expert verified

The pressure attained when the volume is decreased to $0.410 m^{3}$ and the temperature is increased to $157 ° C$ is $19.66 * 10^{3} P a$ .

Step by step solution

Expression for pressure from the ideal gas equation

The ideal gas equation is PV=nRT, whereP is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, T is the temperature.

Write this as an equation for pressure $P = \frac{n R T}{V}$ , modified for initial and final cases as,

$P_{1} = \frac{n R T_{1}}{V_{1}}$

$P_{2} = \frac{n R T_{2}}{V_{2}}$ ., where n and R are constants. Combining both equations $\frac{P_{2}}{P_{1}} = \frac{\frac{T_{2}}{V_{2}}}{\frac{T_{1}}{V_{1}}}$ , solve for the final pressure $P_{2}$ as $P_{2} = \frac{P_{1} T_{2} V_{1}}{T_{1} V_{2}}$ .

Calculating pressure

Substitute the values for $P_{1} = 7.50 * 10^{3} Pa, T_{2} = 157^{\circ} C = 157 + 273 = 430 K$ , $V_{1} = 0.750 m^{3}, V_{2} = 0.410 m^{3}, T_{1} = 27^{\circ} C = 27 + 273 = 300 K in P_{2} = \frac{P_{1} T_{2} V_{1}}{T_{1} V_{2}}$ $P_{2} = \frac{7.50 * 10^{3} * 430 * 0.750}{300 * 0.410} P_{2} = 19.66 * 10^{3} P a$

Hence, the new pressure attained is $19.66 * 10^{3} P a$ .

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Expression for pressure from the ideal gas equation

Calculating pressure

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