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31 votes
An ideal gas in a sealed container has an initial volume of 2.80 L. At constant pressure, it is cooled to 18.00 °C, where its

final volume is 1.75 L. What was the initial temperature?
Ti =
'c

User DeBorges
by
2.5k points

2 Answers

22 votes
22 votes

Final answer:

The initial temperature (Ti) of the gas is 291.15 K.

Step-by-step explanation:

To solve this problem, we can use the relationship between volume and temperature at constant pressure, known as Charles's Law. According to Charles's Law, the volume and temperature of a gas are directly proportional as long as the pressure and amount of gas are constant.

Using the formula V₁/T₁ = V₂/T₂, we can plug in the values given:

  1. Initial volume (V₁) = 2.80 L
  2. Initial temperature (T₁) = ?°C
  3. Final volume (V₂) = 1.75 L
  4. Final temperature (T₂) = 18.00 °C

Converting the temperatures to Kelvin (K), we have:

  • T₁ = 2.80 L * (18.00 °C + 273.15 K) / 2.80 L = 18.00 °C + 273.15 K = 291.15 K
  • T₂ = 1.75 L * (18.00 °C + 273.15 K) / 2.80 L = 18.00 °C + 273.15 K = 291.15 K

Therefore, the initial temperature (Ti) is 291.15 K.

User Thalles Noce
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2.8k points
16 votes
16 votes

Answer:


T_1=-91.18\°C

Step-by-step explanation:

Hello there!

In this case, given the T-V variation, we understand it is possible to apply the Charles' law as shown below:


(T_1)/(V_1)= (T_2)/(V_2)

Thus, since we are interested in the initial temperature, we can solve for T1, plug in the volumes and use T2 in kelvins:


T_1= (T_2V_1)/(V_2)\\\\T_1= ((18.00+273.15)K(1.75L))/((2.80L))\\\\T_1=182K-273.15\\\\T_1=-91.18\°C

Best regards!

User Nihal Sangeeth
by
2.4k points