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Gekata [30.6K]
3 years ago
6

A sample of carbon dioxide gas has a density of g/L at a pressure of 0.889 atm and a temperature of 55.0 °C. Assume ideal behavi

or,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.
Chemistry
1 answer:
gladu [14]3 years ago
5 0

Answer:

163.5 °C

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 686 mL.

Initial temperature (T1) = 55 °C.

Final volume (V2) = 913 mL

Initial pressure (P1) = final pressure (P2) = 0.889 atm

Final temperature (T2) =.?

Next, we shall convert celsius temperature to Kelvin temperature.

This can be done as shown below:

Temperature (K) = Temperature (°C) + 273

T(K) = T (°C) + 273

Initial temperature (T1) = 55 °C.

Initial temperature (T1) = 55 °C + 273

Initial temperature (T1) = 328 K

Next, we shall determine the new temperature of the gas.

Since the pressure is constant, we shall determine the new temperature as follow:

V1/T1 = V2 /T2

Initial volume (V1) = 686 mL.

Final volume (V2) = 913 mL

Initial temperature (T1) = 328 K

Final temperature (T2) =.?

V1/T1 = V2 /T2

686/328 = 913/T2

Cross multiply

686 x T2 = 328 x 913

Divide both side by 686

T2 = (328 x 913)/686

T2 = 436.5 K

Finally, we shall convert Kelvin temperature to celsius temperature.

This can be done as shown below:

Temperature (°C) = Temperature (K) – 273

T (°C) = T(K) – 273

T(K) = 436.5 K

T (°C) = 436.5 – 273

T (°C) = 163.5 °C

Therefore, the temperature of the gas sample is 163.5 °C.

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Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

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7.

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= 151 ÷ 1000 kg

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