Question

3.6 gm of an ideal gas was injected into a bulb of internal volume of 8.21 L at pressure P atm and temperature T-K . The bulb was then placed in a thermostat maintained at (T+15)K . 0.6 gm of the gas was let off to keep the original pressure. . Find P and T if molecular weight of gas is 36.

Note- T denotes temperature in Kelvin and P denotes pressure in atm.​

Answer 1

Ideal gase equation

This question says us that there is an ideal gas of 3.6 gram (g) which is injected into the bulb whose volume is 8.21 liter (l), pressure is P atmosphere (atm) and temperature is T-Kelvin (K). Then the bulb is placed in the thermostat maintained at (T + 15) Kelvin (K), in which 0.6 gram (g) gas is released to maintain the original pressure. We have to find the values Pressure and Temperature.

According to the given question, we have been given that,

• 3.6 gm of an ideal gas
• Internal volume of bulb = 8 L
• Weight of gas = 44
• 0.6 gram gas is released
• Tempreture, T = ?
• Pressure, P = ?

We know that, number of mole = weight/molecular weight. So, Initial moles will be:

$\implies n_1 = \dfrac{3.6}{44}$

When the bulb is placed in the thermostat maintained at (T + 15)K, 0.6 gram gas is released. So, Final moles will be:

$\implies n_2 = \dfrac{3.6 - 0.6}{44} = \dfrac{3}{44}$

Since V and P are constant. Then,

$\implies n_1 T_1 = n_2 T_2$

By substituting the known values in the formula, we obtain the following results:

$\implies \dfrac{3.6}{44} \times T = \dfrac{3}{44} \times (T + 15)$

$\implies 3.6T = 3T + 45$

$\implies 3.6T - 3T = 45$

$\implies 0.6T = 45$

$\implies T = \dfrac{45}{0.6}$

$\implies \boxed{T = 75}$

Now, we know the ideal gas equation, that is,

$\implies PV = nRT$

By substituting the known values in the ideal gas equation, we obtain the following results:

$\implies P \times 8 = \dfrac{3.6}{44} \times 0.0821 \times 75$

$\implies 8P = \dfrac{3.6 \times 0.0821 \times 75}{44}$

$\implies 8P = \dfrac{22.167}{44}$

$\implies P = \dfrac{22.167}{44 \times 8}$

$\implies P = \dfrac{22.167}{352}$

$\implies \boxed{P = 0.062}$

Hence, the values of temperature and pressure are 75 K and 0.0062 atm respectively.

$\rule{300}{2}$

Equation of ideal gase:

The ideal gase equation is given by,

$\longrightarrow PV = nRT$

Where,

• P - pressure measured in atmosphere (atm)
• V - volume measured in Liters (L)
• n - moles of gase present
• R - universal gas constant and is same for all gases, that converts the units. Its value is 0.0821 atm.L/mol.K
• T - temperature measured in Kelvin (T)

The ideal gase is also known as equation of the state. This equation is applicable to any gas which approaches the ideal behaviour.