# A gas weighing 6 kg occupies a volume of 8 m3 at a pressure of 2 × 10 ^ 5 Pa and a temperature of 23 ° C. What volume

**A gas weighing 6 kg occupies a volume of 8 m3 at a pressure of 2 × 10 ^ 5 Pa and a temperature of 23 ° C. What volume will the same gas with a mass of 5 kg occupy at a pressure of 4 × 10 ^ 5 Pa and a temperature of 300 K.**

Given:

m1 = 6 kg

V1 = 8 m3

t1 = 23 оС

p1 = 2 * 10 ^ 5 Pa

m2 = 5 kg

p2 = 4 * 10 ^ 5 Pa

T2 = 300 K

M1 = M2

Find:

V2 =?

Solution:

Let’s convert the initial gas temperature to the absolute temperature on the Kelvin scale.

Absolute gas temperature T = t + 273 K.

Then T1 = 23 oC + 273K = 296 K.

According to the equation of state of an ideal gas (Mendeleev-Clapeyron equation)

p * V = (m / M) * R * T. Where R = 8.31 J / (mol * K) is the universal gas constant.

Let us write the Mendeleev-Clapeyron equation for two states of the gas, taking into account that M1 = M2:

p1 * V1 = (m1 / M) * R * T1 (1)

p2 * V2 = (m2 / M) * R * T2 (2)

Let us divide equation (1) by (2).

We get (p1 * V1) / (p2 * V2) = (m1 * T1) / (m2 * T2).

Whence V2 = (p1 * V1 * m2 * T2) / (p2 * m1 * T1).

Let’s make a calculation.

V2 = (2 * 10 ^ 5 Pa * 8 m3 * 5 kg * 300 K) / (4 * 10 ^ 5 Pa * 6 kg * 296 K) = 3.38 m3.

Answer: the volume of gas will be 3.38 m3.