Question

1 mole gas at standard pressure and at 27°C temperature is heated such that its volume and pressure both become doubled. The final temperature will be

• A. 300 K
• B. 600 K
• C. 900 K
• D. 1200 K

Hint:

For problems involving changes in pressure, volume, and temperature, use the combined gas law, which relates all three variables.

Answer 1

The Correct Option is D

This is a problem involving the ideal gas law. The ideal gas law is
given by:
\[ PV = nRT \] where: \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the gas constant, \( T \) is the temperature in Kelvin. We are given:
Initial pressure \( P_1 \), Initial volume \( V_1 \), Initial temperature \( T_1 = 27^\circ C = 273 + 27 = 300 \, \text{K} \), Final pressure \( P_2 = 2P_1 \), Final volume \( V_2 = 2V_1 \). Using the ideal gas law and the proportionality of the variables for a constant amount of gas, we can use the combined gas law: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Substituting the known values: \[ \frac{P_1 V_1}{300} = \frac{2P_1 \times 2V_1}{T_2} \] Canceling the common terms \( P_1 \) and \( V_1 \): \[ \frac{1}{300} = \frac{4}{T_2} \] Now, solve for \( T_2 \): \[ T_2 = 4 \times 300 = 1200 \, \text{K} \] Thus, the final temperature is 1200 K, which corresponds to (D).