Question

1) At a pressure 48 kPa, the gas in a cylinder has a volume of 15 liters. Assuming temperature remains the same, if the volume of the gas is decreased to 8 liters, what is the new pressure? kPa

2) The gas in a cylinder has a volume of 4 liters at a pressure of 116 kPa. The pressure of the gas is increased to 222 kPa. Assuming the temperature remains constant, what would the new volume be? L

3) If a solid piece of naphthalene is heated and remains at 80°C until it is completely melted, you know that 80ºC is the
A. freezing point of naphthalene.
B. melting point of naphthalene.
C. boiling point of naphthalene.
D. both a and b

Answer 1

Answer on question #62218, Chemistry / General Chemistry

1) At a pressure $48\,\mathrm{kPa}$, the gas in a cylinder has a volume of 15 liters. Assuming temperature remains the same, if the volume of the gas is decreased to 8 liters, what is the new pressure? kPa

Solution:

Since isothermal process T=const, we use the Boyle's law

$$V_1 P_1 = V_2 P_2$$ $$P_2 = \frac{V_1 P_1}{V_2}$$ $$P_2 = \frac{15\,L \times 48\,\mathrm{kPa}}{8\,L} = 90\,\mathrm{kPa}$$

Answer: 90 kPa

2) The gas in a cylinder has a volume of 4 liters at a pressure of $116\,\mathrm{kPa}$. The pressure of the gas is increased to $222\,\mathrm{kPa}$. Assuming the temperature remains constant, what would the new volume be? L

Solution:

Since isothermal process T=const, we use the Boyle's law

$$V_1 P_1 = V_2 P_2$$ $$V_2 = \frac{V_1 P_1}{P_2}$$ $$V_2 = \frac{4\,L \times 116\,\mathrm{kPa}}{222\,\mathrm{kPa}} = 2\,L$$

Answer: 2 L

3) If a solid piece of naphthalene is heated and remains at $80^{\circ}\mathrm{C}$ until it is completely melted, you know that $80^{\circ}\mathrm{C}$ is the

A. freezing point of naphthalene.

B. melting point of naphthalene.

C. boiling point of naphthalene.

D. both a and b

Solution:

This freezing and melting point of naphthalene.

Answer: D. both a and b.

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