Question

1 mole of a monatomic and 2 mole of a diatomic gas are mixes the resulting gas is taken through a process in which

Answer 1

Answer:

ΔQ=ΔU+ΔW (Using First law of thermodynamics)ΔQ=ΔU+ΔW (Using First law of thermodynamics)

dQ=dU+dWdQ=dU+dW

ncdT=ncvdT+PdVncdT=ncvdT+PdV

c=cv+dVdTPn ⋯(i)c=cv+dVdTPn ⋯(i)

Also,Also,

PV=nRT ⋯(ii) and PT2V=k (Constant)⋯(iii)PV=nRT ⋯(ii) and PT2V=k (Constant)⋯(iii)

putting equation(iii) in equation(ii),putting equation(iii) in equation(ii),

kVT2V=nRTkVT2V=nRT

kV2=nRT3kV2=nRT3

Differentiating it we get,Differentiating it we get,

2VkdVdT=3nRT2⋯(iv)2VkdVdT=3nRT2⋯(iv)

putting equation(iv) in equation(i),putting equation(iv) in equation(i),

c=cv+3nRT22VkPnc=cv+3nRT22VkPn

c=cv+(32)(RT2PVk)c=cv+(32)(RT2PVk)

But PT2V=k,But PT2V=k,

c=cv+(32)(RVkVk)c=cv+(32)(RVkVk)

c=R2/5+3R2c=R2/5+3R2

c=4Rc=4R

Answer 2

Answer:ΔQ=ΔU+ΔW (Using First law of thermodynamics)ΔQ=ΔU+ΔW (Using First law of thermodynamics)

dQ=dU+dWdQ=dU+dW

ncdT=ncvdT+PdVncdT=ncvdT+PdV

c=cv+dVdTPn ⋯(i)c=cv+dVdTPn ⋯(i)

Also,Also,

PV=nRT ⋯(ii) and PT2V=k (Constant)⋯(iii)PV=nRT ⋯(ii) and PT2V=k (Constant)⋯(iii)

putting equation(iii) in equation(ii),putting equation(iii) in equation(ii),

kVT2V=nRTkVT2V=nRT

kV2=nRT3kV2=nRT3

Differentiating it we get,Differentiating it we get,

2VkdVdT=3nRT2⋯(iv)2VkdVdT=3nRT2⋯(iv)

putting equation(iv) in equation(i),putting equation(iv) in equation(i),

c=cv+3nRT22VkPnc=cv+3nRT22VkPn

c=cv+(32)(RT2PVk)c=cv+(32)(RT2PVk)

But PT2V=k,But PT2V=k,

c=cv+(32)(RVkVk)c=cv+(32)(RVkVk)

c=R2/5+3R2c=R2/5+3R2

c=4Rc=4R

Explanation: