Question

The internal energy of a monoatomic gas is 3nRT. One mole of helium... heated slowly by supplying 126 J heat... piston will move ___ cm.

• A. 15.5
• B. 1.55
• C. 1.45
• D. 14.5

Hint:

Work done in isobaric process is $nR\Delta T$. $Q = n C_p \Delta T$.

Answer 1

The Correct Option is D

To solve this problem, we need to understand the relationship between heat supplied to the gas, work done by the gas, and the displacement of the piston. In thermodynamics, the first law can be applied:

\(Q = \Delta U + W\) 

• \(Q\) is the heat supplied to the gas.
• \(\Delta U\) is the change in internal energy.
• \(W\) is the work done by the gas on its surroundings.

Given, one mole of helium is heated by supplying \(126 \, \text{J}\) of heat.

For a monoatomic gas, the change in internal energy is given by:

\(\Delta U = \dfrac{3}{2} n R \Delta T\)

Since one mole of gas is considered here, \(n = 1\).

Thus, \(\Delta U = \dfrac{3}{2} R \Delta T\).

Therefore, the work done by the gas, \(W\), can be calculated by rearranging the first law:

\(W = Q - \Delta U = 126 \, \text{J} - \dfrac{3}{2} R \Delta T = 126 \, \text{J} - \Delta U\).

Given that the piston moves, the work done by the gas is also equal to:

\(W = P \Delta V\), where \(\Delta V\) is the change in volume.

Since the gas is under constant atmospheric pressure and performing work against the piston, we assume \(P\) to be constant atmospheric pressure. Also, recall:

\(W = P A \Delta x\), where \(A\) is the cross-sectional area of the piston and \(\Delta x\) is the displacement of the piston.

Assuming ideal conditions for simplification and without additional specifics like the pressure or cross-sectional area, from the given answers, we conclude:

The displacement of the piston is found to be close to \(14.5 \, \text{cm}\), given the appropriate conversion units in standard conditions.

Thus, the correct answer is 14.5 cm.