1). A idea gas undergoes isothermal compression, which reduces its volume by 3.08 dm^{3}. The final pressure and volume of the gas are 6.42 bar and 5.38 dm^{3}, respectively. Calculate the original pressure of the gas in (a) bar, (b) atm.

2). At 0.000°C and 1.000 atm, the mass density of the vapor of a hydrocarbon is 1.97 kg m^{–3}. What is the molecular formula of the hydrocarbon under these conditions? What is the hydrocarbon?

3). Calculate the pressure exerted by 1.0 mol ~~C~~_{6}H_{6} C_{2}H_{6} behaving as (a) an ideal gas, (b) a van der Waals gas when it is confined under the following conditions: (i) at 373.15 K in 22.414 dm^{3}, (ii) at 1000 K in 300 cm^{3}.

4). A certain gas obeys the van der Waals equation with a = 0.580 m^{6} Pa mol^{–2}. Its volume is found to be 3.50 × 10–4 m^{3} mol^{–1} at 273 K and 5.0 MPa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure?

5). In an industrial process, nitrogen is heated to 500 K at a constant volume of 1.000 m^{3}. The gas enters the container at 300 K and 100 atm. The mass of the gas is 92.4 kg. Use the van der Waals equation to determine the approximate pressure of the gas at its working temperature of 500 K. For nitrogen, a = 1.352 dm^{6} atm mol^{-2}, b = 0.0387 dm^{3} mol^{-1}.

6). Suppose that 10.0 mol C_{2}H_{6}(g) is confined to 4.860 dm^{3} at 300 K. Predict the pressure exerted by ethane from a) the ideal gas and b) the van der Waals equations of state. Calculate the compression factor based on these calculations. For ethane, a = 5.507 dm^{6} atm mol^{-2}, b = 0.0651 dm^{3} mol^{-1}

## Comments (2)

## Jackie said

at 2:03 am on Oct 6, 2009

Professor, in problem #3, the gas is C6H6, and the van der Waals coefficients you used in the solution are for C2H6 instead. ???

## Paul Cooper said

at 9:56 am on Oct 6, 2009

It's a typo. The question should read C2H6. Thanks for letting me know.

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