1. Given data

• Initial pressure: $P_1=75\text{ kPa}=7.5\times 10^4\text{ Pa}$
• Initial volume: $V_1=1200\text{ cm}^3=1200\times 10^{-6}\text{ m}^3=1.2\times 10^{-3}\text{ m}^3$
• Final volume: $V_2=150\text{ cm}^3=150\times 10^{-6}\text{ m}^3=1.5\times 10^{-4}\text{ m}^3$
• Gas is monoatomic, so $\gamma=\frac{C_p}{C_v}=\frac{5}{3}$

2. Use adiabatic relation

For an adiabatic process: $P_1V_1^{\gamma}=P_2V_2^{\gamma}$ So, $P_2=P_1\left(\frac{V_1}{V_2}\right)^{\gamma}$

Now, $\frac{V_1}{V_2}=\frac{1200}{150}=8$ Hence, $P_2=75\times 8^{5/3}\text{ kPa}$ Since $8^{5/3}=(2^3)^{5/3}=2^5=32$ therefore, $P_2=75\times 32=2400\text{ kPa}$

3. Work done in adiabatic process

Work done by the gas in an adiabatic process is $W_{\text{by}}=\frac{P_1V_1-P_2V_2}{\gamma-1}$

First compute: $P_1V_1=(7.5\times 10^4)(1.2\times 10^{-3})=90\text{ J}$

Also, $P_2V_2=(2.4\times 10^6)(1.5\times 10^{-4})=360\text{ J}$

Thus, $W_{\text{by}}=\frac{90-360}{\frac{5}{3}-1}$ $=\frac{-270}{\frac{2}{3}}=-405\text{ J}$

Negative sign means the gas is compressed, so work is done on the gas.

Therefore, $W_{\text{on}}=405\text{ J}$

4. Check options

• A: $79\text{ J}$ ❌
• B: $405\text{ J}$ ✅
• C: $4050\text{ J}$ ❌
• D: $9590\text{ J}$ ❌

So the correct option is B.