Solution:
By the first law, \(dQ = dU + dW\). If work is done on the gas, \(dW_{\text{by}} = -dW_{\text{on}}\). Given \(dU = dW_{\text{on}}\), so \(dQ = 0\). This represents an adiabatic process.
The internal energy of an ideal monoatomic gas increases by the same amount as work done on the gas, then:
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