A sample of carbon dioxide gas at a pressure of 1.07 atm and a temperature of 166 °C, occupies a volume of 686 mL. If the gas is heated at constant pressure until its volume is 913 mL, the temperature of the gas sample will be ? °C.

Answer 1

#311^@"C"#

The idea here is that the volume and the temperature of a gas have a direct relationship when the pressure and the number of moles of gas are being kept constant #-># this is known as Charles' Law.

A very important thing to remember is that the temperature of the gas must be expressed in Kelvin. In other words, you must always work with the absolute temperature of a gas.

So, start by converting the temperature of the gas to Kelvin by using

#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + 273.15)))#

You will have

#T = 166^@"C" + 273.15 = "439.15 K"#

The volume of the gas will increase as temperature increases and decrease as temperature decreases. Mathematically, this can be written as

#color(blue)(ul(color(black)(V_1/T_1 = V_2/T_2)))#

Here

Rearrange the equation to solve for #T_2#
#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#

Plug in your values to find

#T_2 = (913 color(red)(cancel(color(black)("mL"))))/(686color(red)(cancel(color(black)("mL")))) * "439.15 K" = "584.47 K"#

Finally, convert this to degrees Celsius

#color(darkgreen)(ul(color(black)(t[""^@"C"] = "584.47 K" - 273.15 = 311^@"C")))#

The answer is rounded to three sig figs.

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Answer 2

To find the final temperature of the gas sample, we can use the combined gas law equation:

[ \frac{{P_1 \cdot V_1}}{{T_1}} = \frac{{P_2 \cdot V_2}}{{T_2}} ]

where: ( P_1 = 1.07 , \text{atm} ) (initial pressure) ( V_1 = 686 , \text{mL} ) (initial volume) ( T_1 = 166 , \text{°C} ) (initial temperature) ( P_2 = 1.07 , \text{atm} ) (final pressure, constant pressure) ( V_2 = 913 , \text{mL} ) (final volume) ( T_2 ) is the final temperature we're solving for.

Rearranging the equation to solve for ( T_2 ), we get:

[ T_2 = \frac{{P_2 \cdot V_2 \cdot T_1}}{{P_1 \cdot V_1}} ]

Plugging in the values, we get:

[ T_2 = \frac{{1.07 , \text{atm} \times 913 , \text{mL} \times 166 , \text{°C}}}{{1.07 , \text{atm} \times 686 , \text{mL}}} ]

[ T_2 \approx 344.7 , \text{°C} ]

So, the temperature of the gas sample will be approximately ( 344.7 , \text{°C} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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