Given 120mL of neon gas at 40 degrees C, what teperature corresponds to a volume that has been decreased by 30mL?

(P1V2)/T1 = (P2V2)/T2.
Use V1 = 120 mL and V2 = 120-30=90 mL
Don't forget to change degrees C to Kelvin.
Kelvin = 273 + oC.

Explain Bot Explain Bot answered
8 months ago

To find the temperature that corresponds to a volume decrease of 30 mL, we can use the ideal gas law equation \(\frac{{P_1 \cdot V_2}}{{T_1}} = \frac{{P_2 \cdot V_2}}{{T_2}}\).

Initial volume, \(V_1 = 120 \ mL\)
Final volume, \(V_2 = V_1 - 30 \ mL = 90 \ mL\)
Initial temperature, \(T_1 = 40 \ degrees \ C\)

First, convert the initial temperature from Celsius to Kelvin by adding 273 to the Celsius value:
\(T_1 = 40 + 273 = 313 \ K\)

Now, substitute the values into the equation and solve for \(T_2\):
\(\frac{{P_1 \cdot V_2}}{{T_1}} = \frac{{P_2 \cdot V_2}}{{T_2}}\)

\(\frac{{P_1 \cdot V_2}}{{T_1}} = \frac{{P_2 \cdot V_2}}{{T_2}}\)

\(\frac{{P_1}}{{T_1}} = \frac{{P_2}}{{T_2}}\)

\(P_2 = P_1 \cdot \frac{{T_2}}{{T_1}}\)

Now, rearrange the equation to solve for \(T_2\):

\(T_2 = \frac{{P_2 \cdot T_1}}{{P_1}}\)

To find \(P_2\), we need more information such as the pressure of the gas. The equation cannot be solved without the value of \(P_2\) or the pressure ratio \(P_2/P_1\).

Once you have the pressure ratio, plug in the values into the equation to find \(T_2\).

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