211k views
1 vote
A sample of helium was compressed at 35 °C from a volume of 0.5 L to 0.25 L where the pressure is 500 mmHg. What was the original pressure?

User Kovge
by
7.4k points

1 Answer

5 votes

Answer:

The ideal gas law is a formula that relates the pressure, volume, temperature, and number of moles of a gas. It can be expressed as:

PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the gas constant, and T is the temperature in Kelvin.

Since the number of moles and the temperature are constant in this problem, we can rearrange the formula to solve for the pressure:

P = (nRT) / V

Plugging in the known values, we get:

P = (nRT) / V

= (n * R * (35 + 273)) / 0.25 L

= (n * R * 308 K) / 0.25 L

We are given that the pressure is 500 mmHg and the volume is 0.25 L at this pressure, so we can set these equal:

500 mmHg = (n * R * 308 K) / 0.25 L

Solving for n, we get:

n = (500 mmHg * 0.25 L) / (R * 308 K)

We know that the pressure and volume of the gas at its original state are what we are trying to solve for. Let's call the original pressure P_0 and the original volume V_0. We can then express the relationship between the original and final state of the gas as:

P_0 * V_0 = n * R * T

Substituting the expression we found for n and the known values, we get:

P_0 * V_0 = ((500 mmHg * 0.25 L) / (R * 308 K)) * R * 308 K

Solving for P_0, we get:

P_0 = (500 mmHg * 0.25 L) / (R * 308 K) * (R * 308 K) / V_0

= (500 mmHg * 0.25 L) / (V_0)

Since V_0 = 0.5 L, we can substitute this value to get:

P_0 = (500 mmHg * 0.25 L) / (0.5 L)

= 500 mmHg / 2

= 250 mmHg

Therefore, the original pressure of the gas was 250 mmHg.

User Shery
by
8.1k points