# 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.

8 months ago

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

Given:
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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