An ideal gas of volume `1.5xx10^-3m^3`and at pressure `1.0xx10^5` Pa is supplied with 70 J of energy. The volume increases to `1.7xx10^-3m^3`, the pressure remaining constant. The internal energy of the gas is

Options

A. increased by 90 J
B. Increased by 70 J
C. Increased by 50 J
D. decreased by 50 J

Answer

To find the change in internal energy of the ideal gas, we can use the first law of thermodynamics, which states:

\[ \Delta U = Q - W \]

where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done by the system.

### Step 1: Identify the given values
- Volume before expansion, \(V_1 = 1.5 \times 10^{-3} \, m^3\)
- Volume after expansion, \(V_2 = 1.7 \times 10^{-3} \, m^3\)
- Pressure, \(P = 1.0 \times 10^5 \, Pa\)
- Heat supplied, \(Q = 70 \, J\)

### Step 2: Calculate the change in volume
\[ \Delta V = V_2 - V_1 = 1.7 \times 10^{-3} \, m^3 - 1.5 \times 10^{-3} \, m^3 = 0.2 \times 10^{-3} \, m^3 \]

### Step 3: Calculate the work done by the gas
The work done \(W\) when the gas expands at constant pressure is given by:
\[ W = P \Delta V \]

Substituting the values:
\[ W = (1.0 \times 10^5 \, Pa) \times (0.2 \times 10^{-3} \, m^3) \]
\[ W = 1.0 \times 10^5 \times 0.2 \times 10^{-3} = 20 \, J \]

### Step 4: Apply the first law of thermodynamics
Now we can substitute the values of \(Q\) and \(W\) into the first law equation:
\[ \Delta U = Q - W \]
\[ \Delta U = 70 \, J - 20 \, J = 50 \, J \]

### Conclusion
The internal energy of the gas increases by \(50 \, J\).

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