Question

a sample of carbon monoxide gas occupies 125 ml at 20.0 °c and 455 mm hg. if the volume of gas is 55.5 ml at 898 mm hg, what is the celsius temperature?

1030 °c

257 °c

18 °c

-16 °c

-207 °c

Explanation:

Step1: Recall Combined Gas Law

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), where \(P\) is pressure, \(V\) is volume, and \(T\) is absolute temperature (in Kelvin). First, convert initial temperature to Kelvin: \(T_1 = 20.0 + 273.15 = 293.15\space K\).

Step2: Rearrange Formula for \(T_2\)

From \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), we get \(T_2=\frac{P_2V_2T_1}{P_1V_1}\).

Step3: Substitute Values

Given \(P_1 = 455\space mm\space Hg\), \(V_1 = 125\space mL\), \(T_1 = 293.15\space K\), \(P_2 = 898\space mm\space Hg\), \(V_2 = 55.5\space mL\). Substitute into the formula: \(T_2=\frac{898\times55.5\times293.15}{455\times125}\).
Calculate numerator: \(898\times55.5 = 49839\); \(49839\times293.15\approx49839\times293 = 49839\times(300 - 7)=14951700 - 348873 = 14602827\).
Denominator: \(455\times125 = 56875\).
Then \(T_2\approx\frac{14602827}{56875}\approx257\space K\) (wait, no, miscalculation. Wait, let's recalculate: \(898\times55.5 = 898\times50 + 898\times5.5 = 44900 + 4939 = 49839\). \(49839\times293.15 = 49839\times293 + 49839\times0.15 = 49839\times(300 - 7)+7475.85 = (14951700 - 348873)+7475.85 = 14602827 + 7475.85 = 14610302.85\). Denominator: \(455\times125 = 56875\). So \(T_2 = 14610302.85\div56875\approx257\space K\)? Wait, no, that can't be. Wait, no, I messed up. Wait, \(T_1\) is 293.15 K. Let's do it step by step:

\(T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{898\space mm\space Hg\times55.5\space mL\times293.15\space K}{455\space mm\space Hg\times125\space mL}\)

First, cancel units: \(mm\space Hg\) and \(mL\) cancel.

Calculate \(\frac{898}{455}\approx1.9736\), \(\frac{55.5}{125}\approx0.444\), then multiply by 293.15: \(1.9736\times0.444\approx0.876\), \(0.876\times293.15\approx256.8\space K\). Then convert to Celsius: \(T_2 (^{\circ}C)= 256.8 - 273.15\approx -16.35^{\circ}C\)? Wait, no, that's wrong. Wait, no, I think I flipped the formula. Wait, combined gas law: \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), so \(T_2 = \frac{P_2V_2T_1}{P_1V_1}\). Wait, but if pressure increases and volume decreases, temperature should change. Wait, initial pressure 455, final 898 (increase), initial volume 125, final 55.5 (decrease). So let's recalculate:

\(P_1 = 455\), \(V_1 = 125\), \(T_1 = 293.15\)

\(P_2 = 898\), \(V_2 = 55.5\)

\(T_2 = \frac{898 \times 55.5 \times 293.15}{455 \times 125}\)

Calculate \(898\div455 \approx 1.9736\)

\(55.5\div125 = 0.444\)

Multiply these two: \(1.9736\times0.444 \approx 0.876\)

Then multiply by 293.15: \(0.876\times293.15 \approx 256.8\space K\)

Convert to Celsius: \(256.8 - 273.15 = -16.35^{\circ}C\), which is approximately \(-16^{\circ}C\). Wait, but the option is -16? Wait, but let's check again. Wait, maybe I made a mistake in the formula. Wait, no, combined gas law is for constant amount of gas. So the formula is correct. Wait, let's do the calculation with more precision.

\(898\times55.5 = 898\times55 + 898\times0.5 = 49390 + 449 = 49839\)

\(455\times125 = 56875\)

\(49839\times293.15 = 49839\times293 + 49839\times0.15 = 49839\times(300 - 7) + 7475.85 = (14951700 - 348873) + 7475.85 = 14602827 + 7475.85 = 14610302.85\)

\(14610302.85\div56875 = 14610302.85\div56875 \approx 256.9\space K\)

\(256.9 - 273.15 = -16.25^{\circ}C\), so approximately \(-16^{\circ}C\). So the correct answer is -16 °C.

Answer:

-16 °C (the option with -16 °C)