JEE Advance - Chemistry (1987 - No. 10)

The value of PV for 5.6 litres of an ideal gas is ________ RT, at NTP.
Answer
0.25

Explanation

The problem posed involves determining the value of PV (pressure-volume product) for a given amount of an ideal gas, specifically 5.6 liters at Normal Temperature and Pressure (NTP). To solve this, we will make use of the ideal gas law, the definition of NTP conditions, and the concept of moles in chemistry.

The ideal gas law is given by:

$$ PV = nRT $$

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 ideal gas constant,
  • T is the temperature of the gas.

Normal Temperature and Pressure (NTP) are defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm. According to the conditions set for NTP, the volume of 1 mole of an ideal gas is 22.4 liters.

To find out how many moles are present in 5.6 liters of gas at NTP, we use the relation between the volume of one mole of gas and the provided volume:

$$ n = \frac{V}{22.4 \text{ liters/mole}} $$

$$ n = \frac{5.6 \text{ liters}}{22.4 \text{ liters/mole}} = 0.25 \text{ moles} $$

Substituting this back into the ideal gas equation:

$$ PV = (0.25 \text{ moles}) \cdot R \cdot T $$

At NTP, since the temperature (T) is taken as 273.15 K, and the pressure is 1 atm, the specific value of R can usually be ignored in this context because we are asked for the expression of PV in terms of RT. Therefore:

$$ PV = 0.25 RT $$

Thus, the value of PV for 5.6 liters of an ideal gas at NTP is 0.25 RT.

Comments (0)

Advertisement