If #1/3 L# of a gas at room temperature exerts a pressure of #35 kPa# on its container, what pressure will the gas exert if the container's volume changes to #7/4 L#?

Question

Adrian Ayala · Accepted Answer

#P=20/3 k Pa# #P_1*V_1=P_2*V_2" Boyle-Mariotte's law"# #1/3*cancel(35)=P*cancel(7)/4# #P=4*5/3# #P=20/3 k Pa#

Xavier Adame · Answer

undefined To solve this problem, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when the temperature remains constant.

According to Boyle's Law:

$ P_1 \times V_1 = P_2 \times V_2 $

Where:
- $ P_1 $ is the initial pressure of the gas
- $ V_1 $ is the initial volume of the gas
- $ P_2 $ is the final pressure of the gas
- $ V_2 $ is the final volume of the gas

Given:
- $ P_1 = 35 \, \text{kPa} $
- $ V_1 = \frac{1}{3} \, \text{L} $
- $ V_2 = \frac{7}{4} \, \text{L} $

We can rearrange Boyle's Law to solve for $ P_2 $:

$ P_2 = \frac{P_1 \times V_1}{V_2} $

Now, plug in the given values:

$ P_2 = \frac{35 \times \frac{1}{3}}{\frac{7}{4}} $

Simplify:

$ P_2 = \frac{35}{3} \times \frac{4}{7} $

$ P_2 = \frac{140}{21} $

$ P_2 = 6.67 \, \text{kPa} $

Therefore, the gas will exert a pressure of $ 6.67 \, \text{kPa} $ when the container's volume changes to $ \frac{7}{4} \, \text{L} $.