Answer
See the detailed answer below.
Work Step by Step
a) We can see that the volume of system 1 is constant while the pressure of system 2 is constant.
Gas 1 undergoes an isochoric process while gas 2 undergoes an isobaric process.
And since the system, gas 1 and gas 2, is isolated, the heat lost from 1 is gained by 2.
$$Q_1+Q_2=0$$
$$n_1C_{\rm V}\Delta T_1+n_2C_{\rm P}\Delta T_2=0$$
The two gases are helium which is a monatomic gas, so
$$\frac{3}{ \color{red}{\bf\not}2}n_1 \color{red}{\bf\not}R(T_f-T_{i1})+ \frac{5}{\color{red}{\bf\not}2}n_2 \color{red}{\bf\not}R(T_f-T_{i2})=0$$
$$ 3n_1 (T_f-T_{i1})+ 5n_2 (T_f-T_{i2})=0$$
$$ 3n_1 T_f-3n_1 T_{i1}+ 5n_2 T_f-5n_2 T_{i2}=0$$
$$ 3n_1 T_f+ 5n_2 T_f=3n_1 T_{i1}+5n_2 T_{i2}$$
$$ (3n_1 + 5n_2) T_f=3n_1 T_{i1}+5n_2 T_{i2}$$
$$ T_f=\dfrac{3n_1 T_{i1}+5n_2 T_{i2}}{3n_1 + 5n_2}$$
$$ T_f=\dfrac{3(0.06)(600)+5(0.03)(300)}{3 (0.06)+ 5(0.03)}$$
$$T_f=\color{red}{\bf 464}\;\rm K$$
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b) The heat transferred from the left side to the right side is heat gained by gas 2, which is given by
$$Q_2=\frac{5}{2}n_2R(T_f-T_{i2})$$
$$Q_2=\frac{5}{2}(0.03)(8.31)(464-300)$$
$$Q_2=\color{red}{\bf 102}\;\rm J$$
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c) We know that gas 2 undergoes an isobaric process, so
$$\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}$$
where $V_f=V_i+\Delta V$
$$\dfrac{V_i}{T_i}=\dfrac{V_i+\Delta V}{T_f}\tag 1$$
Now we need to find $V_1$ and since we know $n$ and $T_i$, we need to find $P_i$ of gas 2.
(In this part we are working on gas 2 only).
Since the piston of gas 2 was initially at rest, the initial pressure of gas 2 is given
$$P_{i}=P_a+P_{piston}=P_a+\dfrac{mg}{A}$$
where $m$ is the mass of the piston and $A$ is its cross-sectional area.
$$P_{i}=(1.013\times 10^5)+\dfrac{(2)(9.8)}{\pi \times 0.05^2}=\bf 1.038\times 10^5\;\rm Pa$$
Thus the initial volume
$$V_i=\dfrac{nRT_i}{P_i}=\dfrac{(0.03)(8.31)(300)}{1.038\times 10^5}$$
$$V_i=\bf 750.52\;\rm cm^3$$
Plugging into (1);
$$\dfrac{750.52}{300}=\dfrac{750.52+\Delta V}{464} $$
Hence,
$$\Delta V=Ah=\dfrac{(750.52)(464)}{300}-750.52=\bf 410.284\;\rm cm^3$$
$$ h= \dfrac{410.284}{A} = \dfrac{410.284}{A} = \dfrac{410.284}{\pi (5)^2} $$
$$h=\color{red}{\bf 5.2}\;\rm cm$$
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d) The fraction of geat that converted to work is given by
$$\dfrac{W}{Q}=\dfrac{\text{ Area under the $P_V$}}{Q}=\dfrac{P\Delta V}{Q}$$
$$\dfrac{W}{Q}=\dfrac{(1.038\times 10^5)(410.284\times 10^{-6})}{102}=\color{red}{\bf 0.42}$$
Hence,
$$\boxed{W=0.42Q}$$
