Gas Law Problem: Pressure and Temperature Relationship

Solving a Gas Law Problem

A gas at 300 K occupies 6.50 L at a pressure of 3.50 atm. What will its pressure be at 250 K if its volume is reduced to 4.80 L?

To solve this problem, we can use the ideal gas law equation:

P1V1 / T1 = P2V2 / T2

First, we need to identify the given values:

  • P1 = 3.50 atm (initial pressure)
  • V1 = 6.50 L (initial volume)
  • T1 = 300 K (initial temperature)
  • V2 = 4.80 L (final volume)
  • T2 = 250 K (final temperature)

Now, plug in the values into the equation to find the final pressure (P2):

P2 = (P1V1 / T1) * (T2 / V2)

Calculating the values:

P2 = (3.50 atm * 6.50 L / 300 K) * (250 K / 4.80 L) = 2.71 atm

Therefore, the pressure of the gas at 250 K with a volume of 4.80 L will be 2.71 atm.

Make sure to double-check your calculations and show all your work to understand the process clearly.

a gas at 300 K occupies 6.50 L at a pressure of 3.50 atm. What will its pressure be at 250 K if its volume is reduced to 4.80 L? Please show work and what law you used. You have the values of P1, V1, T1, T2, and V2. Plug in the numbers and get P2.
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