A vertical cylinder, closed at the bottom end, contains 0.0100 mol of ideal gas. It is fitted at the top with a piston that can move freely. The mass of the piston is 14.0 kg and the initial height of the piston above the bottom of the cylinder is 25 cm. What is the temperature of the gas? $( R = 8.31$ $\mathrm { J } / \mathrm { mol } \cdot \mathrm { K } )$
A) 405 K
B) 3620 K
C) 413 K
D) 500 K
E) 290 K

Question

A vertical cylinder, closed at the bottom end, contains  0.0100  mol of ideal gas. It is fitted at the top with a piston that can move freely. The mass of the piston is  14.0 kg and the initial height of the piston above the bottom of the cylinder is  25 cm. What is the temperature of the gas? $( R = 8.31$ $\mathrm { J } / \mathrm { mol } \cdot \mathrm { K } )$
A)  405 K
B)  3620 K
C)  413 K
D)  500 K
E)  290 K

Quizplus · Accepted Answer

The pressure exerted by the gas can be equated to the pressure exerted by the piston. The pressure due to the piston is given by $P = \frac{F}{A} = \frac{mg}{A}$, where $m$ is the mass of the piston, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $A$ is the cross-sectional area of the piston. However, to find the temperature, we don't need to explicitly calculate the pressure because we can use the ideal gas law in the form $P = \frac{nRT}{V}$, where $n$ is the number of moles of gas, $R$ is the gas constant, $T$ is the temperature in Kelvin, and $V$ is the volume of the gas.Given that the pressure $P$ due to the piston must equal the pressure of the gas, and knowing that the force due to the piston is $mg$, we can set up the equation based on the ideal gas law, rearranged to solve for $T$: $T = \frac{P \cdot V}{n \cdot R}$.The volume $V$ can be considered as the area $A$ times the height $h$, but since $P = \frac{mg}{A}$, the area $A$ cancels out in the equation, leaving us with $T = \frac{mg \cdot h}{n \cdot R}$.Substituting the given values: $m = 14.0 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, $h = 0.25 \, \text{m}$ (since 25 cm = 0.25 m), $n = 0.0100 \, \text{mol}$, and $R = 8.31 \, \text{J/mol} \cdot \text{K}$, we get:$$T = \frac{(14.0 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \cdot (0.25 \, \text{m})}{(0.0100 \, \text{mol}) \cdot (8.31 \, \text{J/mol} \cdot \text{K})}$$$$T = \frac{34.3 \, \text{J}}{0.0831 \, \text{J/K}}$$$$T \approx 413 \, \text{K}$$Therefore, the temperature of the gas is approximately 413 K.

A Vertical Cylinder, Closed at the Bottom End, Contains (R=8.31( R = 8.31(R=8.31

A Vertical Cylinder, Closed at the Bottom End, Contains $( R = 8.31$