f4tbecterk
Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

250 mL of argon gas is held in a flexible vessel shown above. If
the pressure changes to 12.0 atm, what is the new volume of
the gas at 20°C?

Sagot :

Eduard22sly

The new volume of the gas at 20 °C is 205 mL

From the question given above, the following data were obtained:

  • Initial volume (V₁) = 250 mL
  • Initial pressure (P₁) = 10 atm
  • Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
  • New pressure (P₂) = 12 atm
  • New temperature (T₂) = 20 °C = 20 + 273 = 293 K
  • New volume (V₂) =?

With the application of the combine gas equation, we can obtain the new volume of the gas as illustrated below:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \\ \\ \frac{10 \times 250}{298} = \frac{12 \times V_2}{293} \\ \\ \frac{2500}{298} = \frac{12 \times V_2}{293} \\ \\ cross \: multiply \\ \\ 298 \times 12 \times V_2 = 2500 \times 293 \\ \\ divide \: both \: side \: by \: 298 \times 12 \\ \\ V_2 = \frac{2500 \times 293}{298 \times 12} \\ \\ V_2 = 205 \: mL \\ [/tex]

Therefore, the new volume of the gas is 205 mL

Learn more about gas laws: https://brainly.com/question/6844441

View image Eduard22sly
onyebuchinnaji

The new volume of the gas at the given pressure and temperature is 205 mL.

The given parameters:

  • Initial volume of argon, V₁ = 250 mL
  • Initial pressure of gas, P₁ = 10 atm
  • Final pressure of the gas, P₂ = 12 atm
  • Initial temperature of the gas, T₁ = 25 ⁰C = 273 + 25 = 298 K
  • Final temperature of the gas, T₂ = 20 ⁰C = 273 + 20 = 293 K

The new volume of the gas is calculated by applying general gas equation as follows;

[tex]\frac{V_1 P_1}{T_1} = \frac{V_2P_2}{T_2} \\\\V_2 = \frac{V_1P_1 T_2}{T_1P_2} \\\\V_2 = \frac{(250\ mL) \times 10 \times 293}{298 \times 12} \\\\V_2 = 204.84 \ mL\\\\v_2 = 205 \ mL[/tex]

Thus, the new volume of the gas at the given pressure and temperature is 205 mL.

Learn more about general gas equation here:

View image onyebuchinnaji
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.