Question

The heat energy that must be supplied to 14 gram of nitrogen at room temperature to raise its temperature by $48^\circ C$ at constant pressure is (Molecular weight of nitrogen = 28, R = gas constant, $C_p = 7/2 R$ for diatomic gas) ______.

• A. 76 R
• B. 84 R
• C. 90 R
• D. 96 R

Hint:

Always watch out for $C_p$ vs $C_v$! If the container was rigid/sealed (constant volume), you would use $C_v = \frac{5}{2} R$ instead, yielding $60R$. Read the constraint ("constant pressure") carefully!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the amount of heat energy ($Q$) required to raise the temperature of a specific mass of nitrogen gas under constant pressure conditions.

Step 2: Detailed Explanation:
The formula for heat energy exchanged at constant pressure is:
$Q = n \cdot C_p \cdot \Delta T$
where:
$n$ = Number of moles of the gas
$C_p$ = Molar heat capacity at constant pressure
$\Delta T$ = Change in temperature
1. Calculate the number of moles ($n$):
Nitrogen gas exists as diatomic molecules ($N_2$).
We are given the Molecular weight (Molar mass) = $28 \text{ g/mol}$.
Given Mass = $14 \text{ g}$.
$n = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{14}{28} = 0.5 \text{ moles}$.
2. Identify the other variables:
We are given the molar heat capacity $C_p = \frac{7}{2} R$.
We are given the temperature rise $\Delta T = 48^\circ\text{C}$ (which is identical to a rise of 48 K).
3. Calculate the Heat Energy ($Q$):
Substitute all the values into the heat formula:
$Q = (0.5) \times \left( \frac{7}{2} R \right) \times (48)$
$Q = \left( \frac{1}{2} \right) \times \left( \frac{7}{2} R \right) \times (48)$
$Q = \frac{7}{4} R \times 48$
Divide 48 by 4:
$Q = 7 R \times 12$
$Q = 84 R$

Step 3: Final Answer:
The heat energy supplied is 84 R, matching option (b).