Answer
See the detailed answer below.
Work Step by Step
a) The piston is now stationary, and the upper part above the piston is a vacuum, so the pressure of the gas is given by the pressure due to the piston weight.
$$P_{gas}=P_{piston} =\dfrac{m_{piston}g}{A} $$
where the mass of the position is given by $m=\rho V=\rho_{Cu}Ah$
$$P_{ }=\dfrac{\rho_{Cu} \color{red}{\bf\not} Ahg}{ \color{red}{\bf\not} A} $$
where $h$ is the thickness of the piston.
$$P_{ }=\rho_{Cu}hg$$
Plugging the known;
$$P_{ }=(8920)(0.04)(9.8)=$$
$$P_{ }=\color{red}{\bf 3.5\times 10^3}\;\rm Pa$$
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b) Now we need to find the number of moles of the gas so we can find the number of molecules where the number of molecules is given by
$$N=nN_{\rm A}\tag 1$$
$$PV=nRT$$
Hence,
$$n=\dfrac{PV}{RT}$$
The volume of the gas is given by $V=Ah_1=\pi r^2 h_1$
$$n=\dfrac{ \pi r^2 h_1P}{RT}$$
Plugging the known;
$$n={\bf 8.13\times 10^{-4}}\;\rm mol$$
Plugging into (1)
$$N=\left[\dfrac{ \pi r^2 h_1P}{RT}\right]N_{\rm A} $$
Plugging the known;
$$N=\left[\dfrac{ \pi (0.03)^2 (0.20)(3.5\times 10^3)}{(8.31)(20+273)}\right](6.02\times 10^{23})$$
$$N=\color{red}{\bf 4.89\times 10^{20}}\;\rm molecule$$
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c) We know that the pressure of the gas is measured by the weight of the piston and since the weight of the piston is constant and no extra mass added or moved, then the pressure is constant.
Thus, this process is an isobaric process.
Hence,
$$Q=nC_{\rm p}\Delta T$$
$$Q=nC_{\rm p}( T_f-T_i)$$
So,
$$T_f=\dfrac{Q}{nC_{\rm p}}+T_i$$
Plugging the known;
$$T_f=\dfrac{(2)}{( 8.13\times 10^{-4}) (29.1)}+(20+273)=\bf 377\;\rm K$$
$$T_f=\color{red}{\bf104}^\circ\;\rm C$$
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d) For an isobaric process,
$$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$$
Hence,
$$\dfrac{ \color{red}{\bf\not} Ah_1}{T_1}=\dfrac{ \color{red}{\bf\not} Ah_2}{T_2}$$
where $h_2 $ is the gas final height.
$$h_2=\dfrac{h_1T_2}{T_1}$$
Plugging the known;
$$h_2=\dfrac{(20)(377)}{(20+273)}$$
$$h_2=\color{red}{\bf25.7}\;\rm cm$$
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e) We know that the work done on the gas is given by
$$W=-\int PdV$$
where $P$ is constant,
$$W=-P\int_i^f dV=-P(V_f-V_i)$$
$$W= -P(Ah_2-Ah_1)$$
$$W= -PA(h_2-h_1)$$
$$W= -P(\pi r^2)(h_2-h_1)$$
Plugging the known;
$$W= -\pi (3.5\times 10^3)( 0.03^2)(0.257-0.20)$$
$$W=\color{red}{\bf-0.56} \;\rm J$$
