Question

a cylindrical flask is fitted with an airtight piston that is free to slide up and down, as shown in figure below. a mass rests on top of the piston. the initial temperature of the system is 313 k, and the pressure of the gas is held constant at 137 kpa. the temperature is now increased until the height of the piston rises from 23.4 cm to 26.0 cm. what is the final temperature of the gas?

a. 172 k
b. 523 k
c. 935 k
d. 754 k
e. 348 k

Explanation:

Step1: Identify the relevant gas - law

Since the pressure of the gas is held constant, we use Charles's law, which is $\frac{V_1}{T_1}=\frac{V_2}{T_2}$. For a cylinder, $V = Ah$ (where $A$ is the cross - sectional area and $h$ is the height). Since $A$ is constant, we can write $\frac{h_1}{T_1}=\frac{h_2}{T_2}$.

Step2: Rearrange the formula to solve for $T_2$

We can re - arrange $\frac{h_1}{T_1}=\frac{h_2}{T_2}$ to get $T_2=\frac{h_2T_1}{h_1}$.

Step3: Substitute the given values

We are given $h_1 = 23.4$ cm, $T_1=313$ K, and $h_2 = 26.0$ cm. Substituting these values into the formula $T_2=\frac{h_2T_1}{h_1}$, we have $T_2=\frac{26.0\times313}{23.4}$.
$T_2=\frac{8138}{23.4}\approx348$ K.

Answer:

e. 348 K