If #7/5# #L# of a gas at room temperature exerts a pressure of #3# #kPa# on its container, what pressure will the gas exert if the container's volume changes to #5/3# #L#?

Question

Layla Ahmed · Accepted Answer

Start with the Combined Gas Law: #(P_1V_1)/T_1=(P_2V_2)/T_2#

Temperature is constant, so #(P_1V_1)=(P_2V_2) to P_2=(P_1V_1)/V_2 = (3*(7/5))/(5/3) = (21/5)/(5/3)=36/25# #kPa# Boyle, Charles, and Gay-Lussac's Laws—which relate different combinations of pressure, volume, and temperature—could all be committed to memory, but I think it would be easier to just learn the Combined Gas Law and then take away the constant. #(P_1V_1)/T_1=(P_2V_2)/T_2# Since the temperature is maintained constant in this instance, we can disregard it: #(P_1V_1)=(P_2V_2)# Organizing: #P_2=(P_1V_1)/V_2#

Amelia Adams · Answer

undefined Use Boyle's Law to find the new pressure:

$$P_1 \times V_1 = P_2 \times V_2$$

Given:
- Initial pressure ($P_1$) = 3 kPa
- Initial volume ($V_1$) = $7/5$ L
- Final volume ($V_2$) = $5/3$ L

1. Substitute the given values into Boyle's Law equation:
$$3 \text{ kPa} \times \frac{7}{5} \text{ L} = P_2 \times \frac{5}{3} \text{ L}$$

2. Solve for $P_2$:
$$P_2 = \frac{(3 \text{ kPa} \times \frac{7}{5} \text{ L})}{\frac{5}{3} \text{ L}}$$
$$P_2 = \frac{21}{5} \text{ kPa}$$

So, the gas will exert a pressure of $ \frac{21}{5} $ kPa when the container's volume changes to $ \frac{5}{3} $ L.